The Workbook

1. \(u_x = 20\cos 30^\circ = 10\sqrt{3}\) m/s; \(u_y = 20\sin 30^\circ = 10\) m/s.

2. \(T = \dfrac{2u_y}{g} = \dfrac{20}{10} = 2\) s.

3. \(H = \dfrac{u_y^2}{2g} = \dfrac{100}{20} = 5\) m. \(R = u_xT = 10\sqrt{3} \times 2 = 20\sqrt{3} \approx 34.6\) m.

\(u_x = 40\) m/s, \(u_y = 30\) m/s. \(T = \dfrac{2 \times 30}{10} = 6\) s. \(H = \dfrac{900}{20} = 45\) m. \(R = 40 \times 6 = 240\) m.

1. \(u_x = 10\) m/s, \(u_y = 10\sqrt{3}\) m/s. \(H = \dfrac{300}{20} = 15\) m. \(T = 2\sqrt{3}\) s, so \(R = 10 \times 2\sqrt{3} = 20\sqrt{3} \approx 34.6\) m.

2. At the top \(v_y = 0\), so the velocity is \(u_x = 10\) m/s, horizontal.

3. The only force is gravity, so the acceleration is \(10\) m/s\(^2\) downward at every point, including the top.

\(v_x\) stays constant because no force acts horizontally (air resistance is ignored). \(v_y\) changes at the rate \(g\) because gravity acts vertically. It decreases on the way up, is zero at the top and grows on the way down.

\(u_x = u_y = 40 \times \dfrac{1}{\sqrt{2}} = 20\sqrt{2}\) m/s. \(T = \dfrac{2 \times 20\sqrt{2}}{10} = 4\sqrt{2} \approx 5.7\) s. \(H = \dfrac{(20\sqrt{2})^2}{20} = 40\) m. \(R = \dfrac{u^2}{g} = \dfrac{1600}{10} = 160\) m.

\(\dfrac{u^2}{g} = 90\) m.

1. \(90\sin 30^\circ = 45\) m.

2. \(90\sin 90^\circ = 90\) m, the greatest possible.

3. \(90\sin 150^\circ = 45\) m. \(15^\circ\) and \(75^\circ\) add to \(90^\circ\) and give the same range.

\(R_{\max} = \dfrac{u^2}{g} = 100\) gives \(u^2 = 1000\), \(u = 10\sqrt{10} \approx 31.6\) m/s.

At \(45^\circ\): \(H = \dfrac{u^2\sin^2 45^\circ}{2g} = \dfrac{u^2}{4g} = \dfrac{R_{\max}}{4} = 25\) m.

1. \(\sin 2\theta\) is the same for \(30^\circ\) and \(60^\circ\): \(\sin 60^\circ = \sin 120^\circ\). \(R = \dfrac{400 \times \sin 60^\circ}{10} = 20\sqrt{3} \approx 34.6\) m.

2. \(30^\circ\): \(H = \dfrac{(10)^2}{20} = 5\) m. \(60^\circ\): \(H = \dfrac{(10\sqrt{3})^2}{20} = 15\) m. Ratio \(1 : 3\).

3. \(T = \dfrac{2u_y}{g}\): \(2\) s and \(2\sqrt{3} \approx 3.5\) s. Ratio \(1 : \sqrt{3}\). The steeper throw stays up longer.

\(\dfrac{H}{R} = \dfrac{u^2\sin^2\theta/2g}{u^2 \cdot 2\sin\theta\cos\theta/g} = \dfrac{\sin\theta}{4\cos\theta} = \dfrac{\tan\theta}{4}\), so \(\tan\theta = \dfrac{4H}{R}\).

\(R = 4H\): \(\tan\theta = 1\), \(\theta = 45^\circ\). \(R = 4\sqrt{3}H\): \(\tan\theta = \dfrac{1}{\sqrt{3}}\), \(\theta = 30^\circ\).

\(5 = \dfrac{100\sin 2\theta}{10}\) gives \(\sin 2\theta = \dfrac{1}{2}\), so \(2\theta = 30^\circ\) or \(150^\circ\): \(\theta = 15^\circ\) or \(75^\circ\).

\(t = \dfrac{x}{u\cos\theta}\). Put this into \(y\): \(y = u\sin\theta \cdot \dfrac{x}{u\cos\theta} - \dfrac{1}{2}g\dfrac{x^2}{u^2\cos^2\theta} = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}\).

\(y\) is a quadratic in \(x\) with a negative \(x^2\) term: a parabola opening downward.

1. Compare with \(y = x\tan\theta - \ldots\): \(\tan\theta = 1\), so \(\theta = 45^\circ\).

2. \(\dfrac{g}{2u^2\cos^2\theta} = \dfrac{1}{40}\) with \(\cos^2 45^\circ = \dfrac{1}{2}\): \(\dfrac{10}{u^2} = \dfrac{1}{40}\), so \(u^2 = 400\) and \(u = 20\) m/s.

3. \(y = 0\) at \(x = 40\) m, the range. The top is at \(x = 20\) m: \(H = 20 - 10 = 10\) m.

1. At \(x = 10\) m: \(y = 10 - \dfrac{100}{40} = 7.5\) m. This is more than \(5\) m, so the ball clears the wall.

2. \(7.5 - 5 = 2.5\) m.

3. At \(x = 35\) m: \(y = 35 - \dfrac{1225}{40} = 35 - 30.6 = 4.4\) m, less than \(5\) m. The ball hits the wall on its way down.

\(R = \dfrac{2u^2\sin\theta\cos\theta}{g}\), so \(\dfrac{g}{2u^2\cos^2\theta} = \dfrac{\sin\theta}{R\cos\theta} = \dfrac{\tan\theta}{R}\). Then \(y = x\tan\theta - \dfrac{x^2\tan\theta}{R} = x\tan\theta\left(1 - \dfrac{x}{R}\right)\).

Question 12: \(\tan\theta = 1\), \(R = 40\): \(y = x\left(1 - \dfrac{x}{40}\right) = x - \dfrac{x^2}{40}\). It agrees.

1. \(v_x = 40\) m/s always. \(v_y = 30 - 10t\): \(20\), \(0\) and \(-20\) m/s at \(1\), \(3\) and \(5\) s.

2. Both: \(\sqrt{40^2 + 20^2} = \sqrt{2000} = 20\sqrt{5} \approx 44.7\) m/s. The instants are \(2\) s before and after the top, so the ball is at the same height with \(v_y\) reversed.

3. At \(t = 3\) s, the top: \(v_y = 0\) and the speed is \(v_x = 40\) m/s, the least on the whole path.

At the top only \(v_x = u\cos 60^\circ = \dfrac{u}{2}\) remains. On landing at the same level the speed is again \(u\), by symmetry: \(v_x = u\cos\theta\) and \(v_y = -u\sin\theta\).

\(\vec u = (u\cos\theta, u\sin\theta)\) and \(\vec v = (u\cos\theta, u\sin\theta - gt)\). Perpendicular means \(\vec u\cdot\vec v = 0\): \(u^2\cos^2\theta + u\sin\theta(u\sin\theta - gt) = 0\), so \(u^2 = ugt\sin\theta\) and \(t = \dfrac{u}{g\sin\theta}\).

\(t = \dfrac{20}{10 \times \sqrt{3}/2} = \dfrac{4}{\sqrt{3}} \approx 2.3\) s. \(T = \dfrac{2u\sin\theta}{g} = 2\sqrt{3} \approx 3.5\) s, so it happens during the flight.

\(v_x = u\cos\theta\) and \(v_y^2 = u^2\sin^2\theta - 2gh\). So \(v^2 = u^2\cos^2\theta + u^2\sin^2\theta - 2gh = u^2 - 2gh\): the angle drops out.

\(v = \sqrt{625 - 400} = 15\) m/s.

1. \(\vec r = (3\hat i + 4\hat j)(0.5) + \dfrac{1}{2}(-10\hat j)(0.25) = 1.5\hat i + 2\hat j - 1.25\hat j = 1.5\hat i + 0.75\hat j\) m.

2. \(\vec v = 3\hat i + 4\hat j - 5\hat j = 3\hat i - \hat j\) m/s: already moving down.

3. \(y = 4t - 5t^2 = 0\) at \(t = 0.8\) s. Then \(\vec r = 3 \times 0.8\,\hat i = 2.4\hat i\) m: the range.

\(\vec r = \vec r_0 + \vec u t + \dfrac{1}{2}\vec a t^2 = (2\hat i + \hat j) + 8\hat i + 4\hat j = 10\hat i + 5\hat j\) m.

\(\vec v = \vec u + \vec a t = 4\hat i + 4\hat j\) m/s, of magnitude \(4\sqrt{2} \approx 5.7\) m/s.

1. Only the \(\hat k\) component feels gravity: \(z = 20t - 5t^2 = 0\) at \(t = 4\) s.

2. The horizontal components are unchanged: \(\vec r = (6\hat i + 8\hat j) \times 4 = 24\hat i + 32\hat j\) m, of magnitude \(\sqrt{576 + 1024} = 40\) m.

\(u_y = 10\) m/s: \(3.75 = 10t - 5t^2\), so \(t^2 - 2t + 0.75 = 0\) and \((t - 0.5)(t - 1.5) = 0\): \(t = 0.5\) s going up and \(1.5\) s coming down.

\(T = \dfrac{2u\sin\theta}{g}\), \(H = \dfrac{u^2\sin^2\theta}{2g}\) and \(R = \dfrac{u^2\sin 2\theta}{g}\) all have \(g\) in the denominator once. Dividing \(g\) by \(6\) multiplies each of them by \(6\).

\(H = R\) gives \(\dfrac{u^2\sin^2\theta}{2g} = \dfrac{2u^2\sin\theta\cos\theta}{g}\), so \(\sin\theta = 4\cos\theta\) and \(\tan\theta = 4\) (\(\theta \approx 76^\circ\)).

1. Vertically the ball starts from rest: \(20 = 5t^2\), so \(t = 2\) s.

2. \(x = u_xt = 5 \times 2 = 10\) m.

3. \(v_x = 5\) m/s (unchanged) and \(v_y = gt = 20\) m/s. Speed \(= \sqrt{25 + 400} \approx 20.6\) m/s.

1. \(45 = 5t^2\) gives \(t = 3\) s.

2. \(R = 10 \times 3 = 30\) m.

3. \(v_y = 30\) m/s and \(v_x = 10\) m/s: \(\tan\phi = \dfrac{v_y}{v_x} = 3\).

They land together. The horizontal throw adds only a horizontal velocity. The vertical motion of both balls is a fall from rest under \(g\), so both take \(\sqrt{2h/g}\).

\(80 = 5t^2\) gives \(t = 4\) s for both throws. Range \(= ut\): \(60\) m at \(15\) m/s and \(120\) m at \(30\) m/s. The throw speed is horizontal and does not affect the vertical fall, so the time is the same; doubling \(u\) doubles the range.

\(x = ut\) gives \(t = \dfrac{x}{u}\). Then \(y = \dfrac{1}{2}gt^2 = \dfrac{gx^2}{2u^2}\), a parabola. For \(u = 10\) m/s: \(y = \dfrac{10 \times 400}{200} = 20\) m.

1. \(\tan 45^\circ = \dfrac{v_y}{v_x} = 1\), so \(v_y = v_x = 20\) m/s on landing.

2. \(v_y = gt\) gives \(t = 2\) s. \(h = \dfrac{1}{2}gt^2 = 20\) m.

3. \(R = 20 \times 2 = 40\) m.

1. \(u_y = 40\sin 30^\circ = 20\) m/s: \(y = 20t - 5t^2\). The ground is at \(y = -60\): \(t^2 - 4t - 12 = 0\), \((t - 6)(t + 2) = 0\), so \(t = 6\) s.

2. \(u_x = 40\cos 30^\circ = 20\sqrt{3}\) m/s: \(x = 20\sqrt{3} \times 6 = 120\sqrt{3} \approx 208\) m.

Up positive: \(u_y = -20\sin 30^\circ = -10\) m/s. \(-75 = -10t - 5t^2\), so \(t^2 + 2t - 15 = 0\) and \(t = 3\) s. \(x = 20\cos 30^\circ \times 3 = 30\sqrt{3} \approx 52\) m.

1. Above the roof: \(H = \dfrac{u_y^2}{2g} = \dfrac{400}{20} = 20\) m. Above the ground: \(60 + 20 = 80\) m.

2. \(t = \dfrac{u_y}{g} = 2\) s. \(x = 20\sqrt{3} \times 2 = 40\sqrt{3} \approx 69\) m.

3. \(v_x = 20\sqrt{3}\) and \(v_y = 20 - 60 = -40\) m/s: \(v = \sqrt{1200 + 1600} = \sqrt{2800} = 20\sqrt{7} \approx 53\) m/s.

\(v_x = u\cos\theta\) and \(v_y^2 = u^2\sin^2\theta + 2gh\) (the ball ends \(h\) below the start). So \(v^2 = u^2\cos^2\theta + u^2\sin^2\theta + 2gh = u^2 + 2gh\).

Question 9: \(\sqrt{1600 + 1200} = \sqrt{2800}\) m/s, the same.

1. Gravity has a component down the slope and one into the slope: \(a_x = -g\sin\alpha\), \(a_y = -g\cos\alpha\).

2. \(a_x = -10 \times 0.5 = -5\) m/s\(^2\) and \(a_y = -10 \times \dfrac{\sqrt{3}}{2} = -5\sqrt{3} \approx -8.7\) m/s\(^2\).

1. \(u_x = 20\cos 30^\circ = 10\sqrt{3}\) m/s along the incline; \(u_y = 20\sin 30^\circ = 10\) m/s perpendicular to it.

2. \(y = u_yt - \dfrac{1}{2}g\cos\alpha\,t^2 = 0\) gives \(T = \dfrac{2u_y}{g\cos\alpha} = \dfrac{20}{5\sqrt{3}} = \dfrac{4}{\sqrt{3}} \approx 2.3\) s.

3. \(R = u_xT - \dfrac{1}{2}g\sin\alpha\,T^2 = 10\sqrt{3} \times \dfrac{4}{\sqrt{3}} - \dfrac{1}{2} \times 5 \times \dfrac{16}{3} = 40 - \dfrac{40}{3} = \dfrac{80}{3} \approx 26.7\) m.

\(R_{\max} = \dfrac{400}{10(1 + 0.5)} = \dfrac{80}{3}\) m. The angle between the incline and the vertical is \(60^\circ\); half of it is \(30^\circ\) from the incline. This is exactly the throw of Question 12, which is why the two ranges agree: that throw was already the best one.

Perpendicular to the incline: \(y = u\sin\theta\,t - \dfrac{1}{2}g\cos\alpha\,t^2\). Setting \(y = 0\) for \(t \ne 0\): \(T = \dfrac{2u\sin\theta}{g\cos\alpha}\).

With \(\alpha = 0\), \(\cos\alpha = 1\) and \(T = \dfrac{2u\sin\theta}{g}\), the usual time of flight.

The ball is farthest from the incline when \(v_y = 0\): \(t = \dfrac{u_y}{g\cos\alpha} = \dfrac{10}{5\sqrt{3}} = \dfrac{2}{\sqrt{3}} \approx 1.15\) s, half the time of flight.

Distance \(= \dfrac{u_y^2}{2g\cos\alpha} = \dfrac{100}{10\sqrt{3}} = \dfrac{10}{\sqrt{3}} \approx 5.8\) m.

1. The incline slopes down at \(30^\circ\) and the throw is \(30^\circ\) above it, so the throw is horizontal.

2. \(T = \dfrac{2u\sin\theta}{g\cos\alpha} = \dfrac{4}{\sqrt{3}}\) s, the same as before: the perpendicular motion is unchanged.

3. Now gravity's component along the slope helps: \(R = u_xT + \dfrac{1}{2}g\sin\alpha\,T^2 = 40 + \dfrac{40}{3} = \dfrac{160}{3} \approx 53\) m, twice the range up the slope.

Horizontal-vertical axes from the top: \(x = 10t\), \(y = 5t^2\) (down). The ball is on the incline when \(\dfrac{y}{x} = \tan 30^\circ\): \(\dfrac{5t^2}{10t} = \dfrac{1}{\sqrt{3}}\), so \(t = \dfrac{2}{\sqrt{3}} \approx 1.15\) s.

Then \(x = \dfrac{20}{\sqrt{3}}\) m, and the distance along the incline is \(\dfrac{x}{\cos 30^\circ} = \dfrac{20}{\sqrt{3}} \times \dfrac{2}{\sqrt{3}} = \dfrac{40}{3} \approx 13\) m.

\(\dfrac{R_{\text{down}}}{R_{\text{up}}} = \dfrac{1 + \sin\alpha}{1 - \sin\alpha} = \dfrac{1.5}{0.5} = 3\).

\(v_y^2 = 2gh = 400\), so \(v_y = 20\) m/s. \(v = \sqrt{15^2 + 20^2} = 25\) m/s. \(\tan\phi = \dfrac{20}{15} = \dfrac{4}{3}\), so \(\phi = 53^\circ\).

1. \(500 = 5t^2\) gives \(t = 10\) s.

2. \(x = 100 \times 10 = 1000\) m.

3. The packet keeps the plane's horizontal velocity, so the plane is directly above it when it lands.

\(1.25 = 5t^2\) gives \(t = 0.5\) s. Range \(= 10 \times 0.5 = 5\) m.

Yes. Without gravity the bullet would travel along the line of aim and reach the monkey's starting point. Gravity pulls the bullet and the monkey down by the same \(\dfrac{1}{2}gt^2\) in the same time, so they meet below that point.

The incline is already at \(\alpha\) to the horizontal, so the throw is at \(\theta + \alpha\) to the horizontal: \(60^\circ\) in Question 12.

\(u_x = 20\cos 60^\circ = 10\) m/s and \(u_y = 20\sin 60^\circ = 10\sqrt{3}\) m/s. Check: \(\sqrt{100 + 300} = 20\) m/s.

The fall takes \(t = \sqrt{\dfrac{2h}{g}}\), and in that time the ball moves \(R = ut = u\sqrt{\dfrac{2h}{g}}\) horizontally.

\(R = 20\sqrt{\dfrac{90}{10}} = 20 \times 3 = 60\) m.