Question

An object is released from rest at a height of 125m above horizontal ground and falls freely under gravity, hitting a moving target P. This target P is moving in a straight line, with constant acceleration 0.6 m/sec^2.
At the instant the object is released, P passes through a point O with speed 5m/sec. Find the distance from O to the point, where P is hit by object.​

Answer 1

Given :

An object released from rest and freely falling under gravity from a height of :

125 m from horizantal ground

Target P is moving in a straight line with a constant acceleration of :

$a = .6 \frac{m}{s^{2} }$

At the instant when the object is released P passes through a point O with a speed of :

=$5 \frac{m}{s}$

To Find :

The distance from O to the point P, where P is hit by object = ?

Solution :

Since the object is dropped , so its initial velocity is :

u = 0

Now as it covers a distance of 125 m in downward direction , so distance is negative , sinc we we are taking the upward direction as positive :

s = - 125 m

And a =  g = - 10 $\frac{m}{s^{2} }$

Let t be the time it takes to come to the ground , so using 2nd equation of motion we can write :

$S = ut + \frac{1}{2} at^{2}$

So on putting the values we get :

$-125 = -\frac{1}{2}\times 10t^{2}$

t = 5 sec

Now for  the linearly or horizantally moving particle P, since it will take the same time i.e. 5 sec to reach the point of collision , so here again using the 2nd equation of motion and putting the values we get :

$s = 5\times 5 +\frac{1}{2} \times 0.6 \times 5^{2}$

$s = 25 +\frac{15}{2}$

$s= 25 + 7.5$

s = 32.5 m

So the required distance from O to P to reach the point of collision is 32.5 m.