Question

The soccer ball is kicked at 33° from the edge of the building with an initial velocity of 17 m/s and lands 61 meters away from the wall. How tall is the building that the child is standing on?

Answer 1

Answer on Question #54850 – Physics – Mechanics, Kinematics, Dynamics

Question: the soccer ball is kicked at $33{}^{\circ}$ from the edge of the building with an initial velocity of $17\,m/s$ and $61\,m$ away from the wall. How tall is the building that the child is standing on?

Answer:

We have the following situation:

The ball travels along the parabola:

$$y = h + v_{oy}t - \frac{g t^{2}}{2}$$

Motion along the $x$ direction is uniform, therefore the time of fall of the ball is

$$t_{f} = \frac{d}{v_{ox}} = \frac{61}{17 \cdot \cos 33{}^{\circ}} = 4.28\,s$$

At the time $t_f$ the ball is at the point $y = 0$, therefore we can determine $h$ from the first equation:

$$0 = h + v_{oy}t_{f} - \frac{g t_{f}^{2}}{2}$$ $$h = \frac{g t_{f}^{2}}{2} - v_{oy}t_{f} = \frac{9.8 \cdot (4.28)^{2}}{2} - 17 \cdot \sin 33{}^{\circ} \cdot 4.28 = 44.64\,m$$

Finally,

$$h = 44.64\,m$$

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