Question

*Physics Problem#3*
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

Thus the distance from point O to point P is 32.5 m

Explanation:

Give data:

• Height "h" = 125 m
• Acceleration "a" = 0.6 m/s^2
• Speed of object "v" = 5 m /s
• To find: Distance "d" = ?

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 , since we we are taking the upward direction as positive :

s = - 125 m

And a =  g = - 10 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 + 1 / 2 at^2

So on putting the values we get :

-125 = -1 / 2 x  10 t^2  

t = 5 sec

Now for the linearly or

Now using 2nd equation of motion and putting the values we get:

S = 5 x 5 + 1/2 x 0.6 x 5^2

S = 25 + 15 / 2

s = 25 + 7.5

S = 32.5 m

Thus the distance from point O to point P is 32.5 m