Question

A rock is thrown upwards from a building 55.8m high. As the rock falls it barely missed the building and falls towards the ground in a total for 4 seconds from release. Find the initial velocity and maximum height the rock traveled.

Answer 1

Question #15829

First, lets use the law of conservation of energy for point where the rock is thrown upwards and the point, where it has zero velocity (reached maximum height):

$$\frac {m v _ {0} ^ {2}}{2} = m g h ^ {\prime} \Rightarrow h ^ {\prime} = \frac {v _ {0} ^ {2}}{2 g}$$

( $h'$ is the height measured from the building to the maximum height point).

Next, at the point of stop (maximum height): $\nu_{0} = g t^{\prime}$ (2).

Also, let $t$ note the time for moving from the maximum height point (with zero initial velocity) to the ground. Then, $\frac{g t^2}{2} = h + h'$ (h is the height of the building). One might rewrite this as

$h + h' = \frac{g}{2} (t_s - t')^2$ , where $t_s = 4s$ is the summary time. Plugging (1) and (2) into the latter equation gives: $h + \frac{v_0^2}{2g} = \frac{g}{2} (t_s^2 - 2\frac{t_s v_0}{g} + \frac{v_0^2}{g^2})$ , and simplifying, gives:

$$v _ {0} = \frac {1}{t _ {s}} \left(\frac {g t _ {s} ^ {2}}{2} - h\right) = 5.65 \, \mathrm{m/s}.$$

Then, the maximum height is $h_{\max} = h + h' = h + \frac{v_0^2}{2g} \approx 57.43 \, \mathrm{m}$ .