Question

An object is projected upward with initial velocity \({v_0}\) meters per second from a point \({s_0}\) meters above the ground. Show that

\({\left( {v\left( t \right)} \right)^2} = {v_0}^2 - 19.6\left( {s\left( t \right) - {s_0}} \right)\)

Short answer

It is proven that \({\left( {v\left( t \right)} \right)^2} = {v_0}^2 - 19.6\left( {s\left( t \right) - {s_0}} \right)\).

Step-by-step solution

Finding velocity

Acceleration of such a motion is due to the gravity:

So,\(a\left( t \right) = - g = - 9.8m/{s^2}\).

Now, we know that velocity is anti-derivative of acceleration.

So, the general anti-derivative of\(a\left( t \right) = - 9.8m/{s^2}\)is,

\(v\left( t \right) = - 9.8t + {v_0},\;\;\;\;{v_0}\left( {{\rm{initial velocity}}} \right){\rm{ is constant}}\)

Find position function

Here we have,\(v\left( t \right) = - 9.8t + {v_0},\;\;\;\;{v_0}\left( {{\rm{initial velocity}}} \right){\rm{ is constant}}\)

We know that position is anti-derivative of velocity.

So, take anti-derivative of\(v\left( t \right) = - 9.8t + {v_0}\).

\(s\left( t \right) = - 9.8\frac{{{t^2}}}{2} + {v_0}t + {s_0},\;\;\;\;{v_0}\left( {{\rm{initial velocity}}} \right){\rm{,}}{{\rm{s}}_0}\left( {{\rm{initial position}}} \right){\rm{ are any constant}}\).

Prove that \({\left( {v\left( t \right)} \right)^2} = {v_0}^2 - 19.6\left( {s\left( t \right) - {s_0}} \right)\)

Here we have, \(v\left( t \right) = - 9.8t + {v_0},\;\;\;\;{v_0}\left( {{\rm{initial velocity}}} \right){\rm{ is constant}}\)

Now, from above equation,

\(\begin{aligned}{c}{\left( {v\left( t \right)} \right)^2} &= {\left( { - 9.8t + {v_0}} \right)^2}\\ &= \left( { - 9.8t + {v_0}} \right)\left( { - 9.8t + {v_0}} \right)\\ &= 96.04{t^2} - 19.6t + {v_0}^2\\ &= {v_0}^2 - 19.6{v_0}t + 96.04{t^2}\\ &= {v_0}^2 - 19.6\left( { - 4.9{t^2} + {v_0}t} \right)\;\;\;\;\;\;\;\;\left( 1 \right)\end{aligned}\)

Now, we have \(s\left( t \right) = - 9.8\frac{{{t^2}}}{2} + {v_0}t + {s_0}\).

\( \Rightarrow s\left( t \right) - {s_0} = - 4.9{t^2} + {v_0}t\;\;\;\;\;\;\;\left( 2 \right)\)

So, by (1) and (2)

\({\left( {v\left( t \right)} \right)^2} = {v_0}^2 - 19.6\left( {s\left( t \right) - {s_0}} \right)\)

Hence proved.