Question

1. A ball thrown in the x-direction at 4.5m/s from a tower that is 20m tall. Calculatethe speed of the ball when it hits the ground. Also, determine the distance the balllands from the tower.X-direction 4.5mls#1

Answer 1

Given data

*The given speed of the ball in the x-direction is

`u_x=4.5\text{ m/s}`

*The given height is H = 20 m

*The value for the acceleration due to the gravity is

`g=9.8m/s^2`

The formula for the time taken by the ball to hit the ground is given as

`t=\sqrt[]{(2H)/(g)}`

Substitute the known values in the above expression as

`\begin{gathered} t=\sqrt[]{(2*20)/(9.8)} \\ =2.02\text{ s} \end{gathered}`

The formula for the speed of the ball when it hits the ground is given as

`v=\sqrt[]{(u_x)^2+(gt)^2}`

Substitute the known values in the above expression as

`\begin{gathered} v=\sqrt[]{(4.5)^2+(9.8*2.02)^2} \\ =20.3\text{ m/s} \end{gathered}`

Hence, the speed of the ball when it hits the ground is v = 20.3 m/s

The formula for the distance of the ball lands from the tower is given as

`R=u_x* t`

Substitute the known values in the above exprssion as

`\begin{gathered} R=(4.5)*2.02 \\ =9.09\text{ m} \end{gathered}`

Hence, the distance of the ball lands from the tower is R = 9.09 m