Question

a cricket ball of mass 0.3 kilogram is thrown vertically upward with a velocity 14.7 metre per second calculate the kinetic energy and potential energy of ball after 1 seconds​

Answer 1

We have been given that,

Mass of the ball = 0.3 Kg

Initial Velocity = 14.7 m/sec

And we have to find the Kinetic and Potential Energy of the body after 1 second.

After one second,

Velocity of the body would be:-

v = u + at

v = 14.7 + (-10)(1)

v = 4.7 m/s

So, the Kinetic Energy would be:-

KE = $\dfrac{1}{2} mv^{2}$

KE = $\dfrac{1 \times 3 \times 4.7^{2}}{2 \times 10}$

KE = $\dfrac{66.27}{20}$

KE = 3.3135 Joules

Height reached by the body after 1 second would be:-

S = $ut + \dfrac{1}{2} t^{2}$

S = $(14.7)(1) + \dfrac{1}{2} (-10) (1^{2})$

S = 14.7 - 5

S = 9.7 m

So, the potential energy would be:-

PE = mgh

PE = 0.3 x 10 x 9.7

PE = 29.1 Joules

So, the Kinetic Energy and Potential Energy of the ball after 1 second would be 3.3125 and 29.1 Joules respectively !