A ball is thrown vertically upward with a speed of 360 ms a

A ball is thrown vertically upward with a speed of 36.0 m/s.

(a) How high does it rise?
  
(b) How long does it take to reach its highest point?
  
(c) How long does the ball take to hit the ground after it reaches its highest point?
  

(d) What is its velocity when it returns to the level from which it started?

Solution

For a body projected upwards with velocity say U m/s, we have the following:

1.) It will continue to rise with a deceleration of g m/s^2 until its velocity decreases to zero and then again it will start dropping back.

2.) The sum of KE and PE will remain constant throughout the journey of the body.

Now, we try and solve the questions:

a.) We know the relation: v^2 = u^2 - 2gH

Hence, 0 = 36^2 - 2*9.81*H

Hence, H = 66.055 metres

b.) Again, we have the relation: V = U + at

That is, 0 = 36 - 9.81 * t

Hence, time taken to reach the highest point = t = 3.6697 seconds

c.) The ball will again take the same time to get back to the point of launch

hence, the time required = 3.6697 seconds

d.) As the energy needs to be conserved at all points of the travel.

The magnitude of the velocity will be same as before, i.e, 36 m/s , however, the direction will be downwards.