Question

A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hits the ground. (a) find the velocity of the object when it is 30.0 m above the ground. (b) find the total distance the object travels during the fall.

Answer 1

We use the kinematic equation for vertical motion,

$s=ut+\frac{1}{2}gt^2$

Here, $s=-30\ m$, $t=1.50\ s$ and $g=-9.8\ m/s^2$.

Therefore, the initial velocity when the object is at height 30 m, from above equation

$-30\ m= u(1.50\ s)+\frac{1}{2} (-9.8\ m/s^2)(1.50\ s)^2\\\\ u=\frac{-18.98\ m }{1.50\ s} =-12.65\ m/s$

Now the displacement before the reaching the height 30.0 m above the ground, we use kinematic equation

$u^2=U^2+2gh$

Here, $u=-12.65\ m/s$,$U=0$ because freely falling object, released from rest and $g= -9.8\ m/s^2$ and we take height -h because downward falling.

Substituting these values in above equation, we get

$(-12.65\ m/s)^2=0+2(-9.8\ m/s^2)(-h)\\\\ h=\frac{160.023}{19.6 }=8.16\ m$.

Therefore, the total distance the object travels during the fall,  s+h= 30 m+ 8.16 =38.16 m.