Question

You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 49° from the edge of the building with an initial velocity of 13 m/s and lands 65 meters away from the wall. How tall is the building that the child is standing on?

Answer 1

Answer on Question #47597-Physics-Mechanics-Kinematics-Dynamics

You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at $\alpha = 49{}^{\circ}$ from the edge of the building with an initial velocity of $v_0 = 13\frac{m}{s}$ and lands $s = 65$ meters away from the wall. How tall is the building that the child is standing on?

Solution

Ignoring air resistance:

$$V_{x0} = V_0 \cos \alpha = 13 \frac{m}{s} \cos 49 = 8.5 \frac{m}{s}$$ $$V_{y0} = V_0 \sin \alpha = 13 \frac{m}{s} \sin 49 = 9.8 \frac{m}{s}.$$

So how long does it take the ball to travel 65 meters along the $x$ axis?

$$s = V_{x0} t \rightarrow t = \frac{s}{V_{x0}} = \frac{65}{8.5} = 7.6 \, s.$$

We now know the total flight time of the ball was 7.6 seconds.

What about it's height?

$$S_y - S_{y0} = V_{y0} t - \left(\frac{1}{2}\right) g t^2.$$

where $S_{y0}$ is the initial distance up the $y$ axis (ie the height of the roof that we want to find); $S_y = 0$ is the final height of the ball (i.e. ground level); $V_{y0}$ is the initial velocity along the $y$ axis; $g = 9.8 \frac{m}{s^2}$ is the acceleration of gravity; $t$ is the total flight time.

Then

$$- S_{y0} = V_{y0} t - \left(\frac{1}{2}\right) g t^2 \rightarrow S_{y0} = \left(\frac{1}{2}\right) g t^2 - V_{y0} t = \left(\frac{1}{2}\right) 9.8 \cdot 7.6^2 - 9.8 \cdot 7.6 = 208.5 \, m.$$

So the child is on a roof that is 208.5 meters tall.

Answer: 208.5 m.

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