Question

a playground is on the flat roof of a city school, (h_0 = 5.20) m above the street below (see figure). the vertical wall of the building is (h = 6.70) m high, to form a 1.5 - m - high railing around the playground. a ball has fallen to the street below, and a passerby returns it by launching it at an angle of (\theta=53.0^{circ}) above the horizontal at a point (d = 24.0) m from the base of the building wall. the ball takes 2.20 s to reach a point vertically above the wall.
(a) find the speed at which the ball was launched.
(b) find the vertical distance by which the ball clears the wall.
(c) find the horizontal distance from the wall to the point on the roof where the ball lands.

Explanation:

Step1: Analyze horizontal motion

The horizontal - motion of a projectile is a uniform - motion with the equation $x = v_{0x}t$, where $x = d$, $v_{0x}=v_0\cos\theta$, and $t$ is the time of flight to the wall. We can express $v_0$ from this equation. Given $d = 24.0$ m, $\theta = 53.0^{\circ}$, and $t = 2.20$ s. Since $x = v_0\cos\theta\times t$, then $v_0=\frac{d}{t\cos\theta}$.
$v_0=\frac{24.0}{2.20\times\cos(53.0^{\circ})}$
$v_0=\frac{24.0}{2.20\times0.602}$
$v_0=\frac{24.0}{1.3244}\approx18.1$ m/s

Step2: Analyze vertical motion to find the height at $t = 2.20$ s

The vertical - motion of a projectile is a uniformly - accelerated motion with the equation $y = v_{0y}t-\frac{1}{2}gt^{2}$, where $v_{0y}=v_0\sin\theta$, $g = 9.8$ m/s². First, $v_{0y}=18.1\times\sin(53.0^{\circ})=18.1\times0.799\approx14.5$ m/s. Then $y = 14.5\times2.20-\frac{1}{2}\times9.8\times(2.20)^{2}$
$y = 31.9-23.716 = 8.184$ m. The height of the wall is $h = 6.70$ m. The vertical distance by which the ball clears the wall is $\Delta y=y - h$.
$\Delta y=8.184 - 6.70=1.484\approx1.48$ m

Step3: Analyze vertical motion to find the time of flight to the roof

The height of the roof is $h_0 = 5.20$ m. The vertical - displacement from the ground to the roof is $y_1$. The initial vertical velocity $v_{0y}=14.5$ m/s. Using the equation $y_1 = v_{0y}t_1-\frac{1}{2}gt_1^{2}$, where $y_1 = 5.20$ m. So, $5.20=14.5t_1 - 4.9t_1^{2}$, or $4.9t_1^{2}-14.5t_1 + 5.20 = 0$. Using the quadratic formula $t_1=\frac{14.5\pm\sqrt{14.5^{2}-4\times4.9\times5.20}}{2\times4.9}=\frac{14.5\pm\sqrt{210.25 - 101.92}}{9.8}=\frac{14.5\pm\sqrt{108.33}}{9.8}=\frac{14.5\pm10.41}{9.8}$. We get two solutions for $t_1$: $t_{11}=\frac{14.5 + 10.41}{9.8}\approx2.54$ s and $t_{12}=\frac{14.5 - 10.41}{9.8}\approx0.42$ s. The first solution corresponds to the ball's upward - then - downward motion to the roof, and the second solution corresponds to the ball's upward motion before reaching the maximum height.
The horizontal distance from the wall to the point on the roof is $x_2=v_{0x}(t_1 - t)$. $v_{0x}=v_0\cos\theta=18.1\times0.602 = 10.9$ m/s. $t_1 = 2.54$ s, $t = 2.20$ s.
$x_2=10.9\times(2.54 - 2.20)=10.9\times0.34 = 3.706\approx3.71$ m

Answer:

(a) 18.1 m/s
(b) 1.48 m
(c) 3.71 m