Question

An m = 6 kg ball is projected with initial horizontal speed v = 60 m/s from a platform h = 80 meters high.(a) How much time does it take for the ball to reach the ground?(b) What is the ball's kinetic energy when it reaches the ground?

Answer 1

Given the mass of the ball, m= 6 kg, and the horizontal speed, v = 60 m/s.

Also, the height of the platform, h = 80 m

(a) To find the time required to reach the ground, t

Time can be calculated from the formula

`\begin{gathered} h=(1)/(2)gt^2 \\ t=\sqrt[]{(2h)/(g)} \end{gathered}`

Here, g= 9.8 m/s^2 which is the acceleration due to gravity.

Substituting the values in the above equation, we get

`\begin{gathered} t=\sqrt[]{(2*80)/(9.8)} \\ =4.03\text{ seconds} \end{gathered}`

(b) To find the Kinetic energy of the ball.

In order to find kinetic energy, first, we need to find the velocity.

The velocity at the ground, u can be obtained by the formula

`u=v+gt`

Substituting the values, we get

`\begin{gathered} u=60+(9.8*4.03)_{} \\ u=99.49\text{ m/s} \end{gathered}`

Substituting the value of velocity in the below formula of kinetic energy, we get

`\begin{gathered} KE=(1)/(2)mu^2 \\ =(1)/(2)*6*(99.49)^2 \\ =29694.78\text{ J} \end{gathered}`

Thus, the time taken is 4.03 seconds and kinetic energy is 29694.78 J