Question

An m = 8 kg ball is projected with initial horizontal speed v = 40 m/s from a platform h = 80 meters high.(a) How much time does it take for the ball to reach the ground? s(b) What is the ball's kinetic energy when it reaches the ground? J

Answer 1

Given,

The mass of the ball, m=8 kg.

The initial horizontal speed of the ball, v=40 m/s

The height of the platform, h= 80 m

As the ball is projected horizontally, the vertical component of the initial velocity will be zero. The ball will be having only horizontal velocity.

(a)

From the equation of the motion,

$h=v_yt+\frac{1}{2}gt^2$

Where vy is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time that the ball takes to reach the ground.

On substituting the known values,

$\begin{gathered} 80=0+\frac{1}{2}\times9.8\times t^2 \\ \Rightarrow t=\sqrt[]{\frac{80\times2}{9.8}} \\ =4.04\text{ s} \end{gathered}$

Thus the ball will take 4.04 s to reach the ground.

The y-component of the final velocity is given by,

$u_y=v_y+gt$

On substituting the known values,

$\begin{gathered} u_y=0+9.8\times4.04 \\ =39.6\text{ m/s} \end{gathered}$

As there is no acceleration on the ball in the horizontal direction, the horizontal component of the velocity will remain the same. That is,

$u_x=v_x=40\text{ m/s}$

Thus the magnitude of the final velocity is given by,

$u=\sqrt[]{u^2_x+u^2_y}$

On substituting the known values,

$\begin{gathered} u=\sqrt[]{39.6^2+40^2} \\ =56.3\text{ m/s} \end{gathered}$

The kinetic energy is given by,

$K=\frac{1}{2}mu^2$

On substituting the known values,

$\begin{gathered} K=\frac{1}{2}\times8\times56.3^2 \\ =12.68\times10^3\text{ J} \\ =12.68\text{ kJ} \end{gathered}$

Therefore the kinetic energy of the ball when it reaches the ground is 12680 J