A rock is thrown vertically upward from ground level at time $\mathbf{t}\mathbf{=}\mathbf{0}$ . Atrole="math" localid="1656149217888" $\mathbf{t}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\hspace{0.33em}}\mathbf{s}$ ,it passes the top of a tall tower, and 1.0 s later, it reaches its maximum height. What is the height of the tower?

The time taken by the rock to reach the top of the tower is$\mathrm{t}=1.5\mathrm{s}$ .

The time taken by the rock to achieve the maximum height is${\mathrm{t}}^{\text{'}}=2.5\mathrm{s}$ .

With the help of the given time interval, using the kinematics equation the height of the tower can be determined.

The kinematic equations that can be used to calculate the height are,

${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{i}}+\mathrm{at}\left(\mathrm{i}\right)$

$\mathrm{\Delta y}={\mathrm{v}}_{\mathrm{i}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2\left(\mathrm{ii}\right)$

At maximum height velocity of the projectile is zero. Substitute the values in equation (i).

$$\begin{gathered} 0={\mathrm{v}}_{\mathrm{i}}+(-9.8)(2.5) \\ {\mathrm{v}}_{\mathrm{i}}=24.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the initial velocity is 24.5 m/s .

Using the initial velocity and other given values in equation (ii), we can find the height of the tower. Substitute the values in equation (ii).

$$\begin{gathered} ∆\mathrm{y}=(24.5\mathrm{m}/\mathrm{s})(1.5\mathrm{s})+\frac{1}{2}(-9.8\mathrm{m}/{\mathrm{s}}^2){(1.5\mathrm{s})}^2 \\ =25.72\mathrm{m} \\ \approx 26\mathrm{m} \end{gathered}$$

Therefore, the height of the tower is 26 m .