Question

A boy is standing on a road , car throws a ball straight upwards . the car is moving with acceletration 1 m/s^2 ant the projectile velocity is 9.8 m/s .how far behind the boy will ball fall on the car ?

Answer 1

Task. A boy stands on a road in and throws a ball straight upwards. The car is moving with acceleration $a=1\ m/s^{2}$ and the projectile velocity is $v_{0}=9.8\ m/s$. How far behind the boy will ball fall on the car?

Solution. Notice that there is a graviation force acting on the ball, so it will move with constant acceleration $g=9.8\ m/s$ directed downward. Hence the velocity $v_{b}(t)$ and the height $h_{b}(t)$ of the ball at time $t$ is given by the formula:

$v_{b}(t)=v_{0}-gt,\qquad h_{b}(t)=v_{0}t-\frac{gt^{2}}{2}.$

On the other hand, the car is moved with zero initial velocity and acceleration $a$, so its position $d(t)$ at time $t$ is equal to

$d(t)=\frac{at^{2}}{2}.$

We should find $d(\bar{t})$, where $\bar{t}$ is the time when the ball reach the ground, so when $h_{b}(t)=0$.

Let us solve the latter equation:

$h_{b}(\bar{t})=v_{0}\bar{t}-\frac{g\bar{t}^{2}}{2}=0$

It follows that either $t=0$, or

$v_{0}=\frac{g\bar{t}}{2}\qquad\Rightarrow\qquad\bar{t}=\frac{2v_{0}}{g}.$

Substituting values of $v_{0}$ and $g$ we obtain:

$\bar{t}=\frac{2*9.8}{9.8}=2\ s.$

Hence

$d(\bar{t})=d(2\ s)=\frac{a\bar{t}^{2}}{2}=\frac{1*2^{2}}{2}=2\ m.$

Answer. $2m$