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.
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 \noindent 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.
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The only meaningful solution is \(t=5\). The rock is in the air for \(5\) seconds.