A rock is thrown straight up with an initial velocity of $19.6\;{\rm m}/{\rm s}$ from the roof of a building $24.5\;{\rm m}$ above ground level, as shown in Figure 64.

Figure 64

Solution To answer the questions, we need to find the velocity $v=v(t)$ and the distance $s=s(t)$ of the rock as functions of time. We begin measuring time when the rock is released. If $s$ is the distance, in meters, of the rock from the ground, then since the rock is released at a height of $24.5\;{\rm m}$, $s_{0}=s(0)=24.5\;{\rm m}$.

The initial velocity of the rock is given as $v_{0}=v(0)=19.6\;{\rm m}/{\rm s}$. If air resistance is ignored, the only force acting on the rock is gravity. Since the acceleration due to gravity is $-9.8\;{\rm m}/{\rm s}^{2}$, the acceleration $a$ of the rock is $$a=\dfrac{{d\kern-1ptv}}{dt}=-9.8$$

Solving the differential equation, we get $$v(t)=-9.8t+v_{0}$$

Using the initial condition, $v_{0}=v( 0) =19.6\;{\rm m}/{\rm s}$, the velocity of the rock at any time $t$ is $$v(t)=-9.8t+19.6$$

Now we solve the differential equation $\dfrac{ds}{dt}=v(t) =-9.8t+19.6$. Then $${s}(t)={-4.9t^{2}+19.6t+s_{0}}$$

Using the initial condition, $s( 0) =24.5\;{\rm m}$, the distance $s$ of the rock from the ground at any time $t$ is $$s(t)=-4.9t^{2}+19.6t+24.5$$

Now we can answer the questions.

336

The only meaningful solution is $t=5$. The rock is in the air for $5$ seconds.