A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building $\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{s}$after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground$\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{s}$after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

The time taken by the rock to reach maximum height is$1.60s$ .

The time taken by the rock to strike the ground after the launch is $6.00s$.

Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion.The equations which are used in the study are known as kinematic equations of motion.

The firstkinematics equation of motion is written as,

v$v_f=v_0+at$

Now, it is given that time to reach the maximum height is $1.6s$. The final velocity is zero as the rock will come to stop at maximum height. The acceleration is the acceleration due to gravity, but it acts in the downward direction, opposite to the direction of motion.

$\begin{array}{l}0=v_a+(9.8m/s^2\times 1.6s)\\ v_0=9.8\times 1.6\\ =15.68\\ \approx 15.7m/s\end{array}$

Hence, the initial velocity is $15.7\mathrm{m}/\mathrm{s}$.

To find the maximum height, use another kinematic equation.

$∆y=v_{0}t+\frac{1}{2}a{t}^2$

Initial velocity is calculated in step 1. Use the time required to reach the maximum height.

$$\begin{gathered} ∆y=\left(15.68m/s\times 1.6s\right)+\left(\frac{1}{2}\times -9.8m/s^2\times {\left(1.6s\right)}^2\right) \\ =25.088-12.544 \\ =12.544m \\ \approx 12.5m \end{gathered}$$

So, the maximum height reached by the rock is $12.5m$.

From step 3, the initial velocity is$15.68m/s$. The time the rock takes to reach the bottom is$6\sec$. The acceleration due to gravity is in the downward direction. Use a kinematic equation to calculate the height as,

$y-y_0=v_{0}t+\frac{1}{2}a{t}^2$

The height at the ground is zero. The height of the building is $y_0$Substitute the given values.

$$\begin{gathered} 0-y_0=\left(15.68m/s\times 6s\right)+\left(\frac{1}{2}\times -9.8m/s^2\times {\left(6s\right)}^2\right) \\ -y_0=94.08-176.4 \\ {y}_0=82.32 \\ \approx 82.3m \end{gathered}$$

From the above calculations,you can say that building is $82.3m$tall.