Question

A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown? What is the potential energy of the ball when it gets to its maximum height just before falling back to the ground?

Answer 1

The initial velocity is u = 20 m/s

We can use the following equations of motion to solve this question:

v = u + at

s = ut + 1/2 at^2

v^2 = u^2 + 2as

Here, 'u' is the initial velocity, 'v' is the final velocity, 't' is the time taken, 'a' is the acceleration and 's' is the distance traveled. 

In the case of vertical motion (such as this one), acceleration due to gravity is the only acceleration. For simplicity, let us use a value of 10 m/s^2 for it.

a) Maximum height corresponds to a velocity of 0 m/s and after this instant the ball falls down.

Thus, 0^2 = 20^2 + 2(-10)s

Negative acceleration is used to denote the downward direction of acceleration, while the ball is traveling upwards.

Simplifying further, s = 20^2 /(2 x 10) = 20 m.

Thus, the ball will attain a maximum height of 20 m.

b) For the time taken to reach the maximum height:

v = u + at

0 = 20 +(-10) t

So, t = 2 s 

c) Since, the motion of the ball is symmetrical, it will take 2 seconds to rise upwards and 2 seconds to fall back to the ground. Thus, the total journey takes 4 s. (you can also check it by using 20 = 1/2 (10) t^2, for the downward motion of ball).

The velocity at the time of reaching the ground is:

v = u + at

v = 0 + (-10) x 2 = 20 m/s (in the downwards direction).

Hope this helps.

                                

Answer 2

Answer:

Both of the answers are 225

Step-by-step explanation: