Question

a playground is on the flat roof of a city school, $h_{b}=5.20$ 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. the ball takes 2.20 s to reach a point vertically above the wall. (a) find the speed at which the ball was launched. m/s (b) find the vertical distance by which the ball clears the wall. m (c) find the horizontal distance from the wall to the point on the roof where the ball lands. m

Explanation:

Step1: Analyze horizontal motion

The horizontal - motion of a projectile is a uniform - motion with the formula $x = v_{0x}t$, where $x = d$, $v_{0x}=v_0\cos\theta$, and $t$ is the time of flight to the point vertically above the wall. We know that $d = 24.0$ m, $t = 2.20$ s, and $\theta = 53.0^{\circ}$. From $x = v_{0x}t=v_0\cos\theta t$, we can solve for $v_0$.
$v_0=\frac{d}{t\cos\theta}$

Step2: Calculate the initial speed $v_0$

Substitute $d = 24.0$ m, $t = 2.20$ s, and $\theta = 53.0^{\circ}$ ($\cos\theta=\cos53.0^{\circ}=0.602$) into the formula:
$v_0=\frac{24.0}{2.20\times0.602}=\frac{24.0}{1.324}\approx18.1$ m/s

Step3: 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 formula $y = v_{0y}t-\frac{1}{2}gt^{2}$, where $v_{0y}=v_0\sin\theta$, $g = 9.8$ m/s². First, find $v_{0y}=18.1\times\sin53.0^{\circ}=18.1\times0.799 = 14.46$ m/s. Then, $y = 14.46\times2.20-\frac{1}{2}\times9.8\times(2.20)^{2}=31.812 - 23.716=8.096$ m. The height of the wall is $h = 6.70$ m. The vertical distance by which the ball clears the wall is $\Delta y=y - h$.
$\Delta y=8.096 - 6.70 = 1.396\approx1.40$ m

Step4: Analyze the total vertical displacement to find the total time of flight

The total vertical displacement when the ball lands on the roof is $y_{total}=h_b=5.20$ m. Using the equation $y_{total}=v_{0y}t_{total}-\frac{1}{2}gt_{total}^{2}$, where $v_{0y}=14.46$ m/s. So, $5.20 = 14.46t_{total}-4.9t_{total}^{2}$, or $4.9t_{total}^{2}-14.46t_{total}+5.20 = 0$. Using the quadratic formula $t_{total}=\frac{14.46\pm\sqrt{14.46^{2}-4\times4.9\times5.20}}{2\times4.9}=\frac{14.46\pm\sqrt{209.09 - 101.92}}{9.8}=\frac{14.46\pm\sqrt{107.17}}{9.8}=\frac{14.46\pm10.35}{9.8}$. We get two solutions for $t_{total}$: $t_{1}=\frac{14.46 + 10.35}{9.8}=2.53$ s and $t_{2}=\frac{14.46 - 10.35}{9.8}=0.42$ s. The smaller value corresponds to the ball's upward - motion phase, and the larger value $t_{total}=2.53$ s is the total time of flight.
The horizontal distance traveled in time $t_{total}$ is $x_{total}=v_0\cos\theta t_{total}$. The horizontal distance traveled in $t = 2.20$ s is $d = 24.0$ m. The horizontal distance from the wall to the landing point on the roof is $x'=v_0\cos\theta(t_{total}-t)$. Substitute $v_0 = 18.1$ m/s, $\cos\theta = 0.602$, $t_{total}=2.53$ s, and $t = 2.20$ s: $x'=18.1\times0.602\times(2.53 - 2.20)=18.1\times0.602\times0.33 = 3.60$ m

Answer:

(a) 18.1 m/s
(b) 1.40 m
(c) 3.60 m