Given data
*The given speed of the ball in the x-direction is
[tex]u_x=4.5\text{ m/s}[/tex]
*The given height is H = 20 m
*The value for the acceleration due to the gravity is
[tex]g=9.8m/s^2[/tex]
The formula for the time taken by the ball to hit the ground is given as
[tex]t=\sqrt[]{\frac{2H}{g}}[/tex]
Substitute the known values in the above expression as
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times20}{9.8}} \\ =2.02\text{ s} \end{gathered}[/tex]
The formula for the speed of the ball when it hits the ground is given as
[tex]v=\sqrt[]{(u_x)^2+(gt)^2}[/tex]
Substitute the known values in the above expression as
[tex]\begin{gathered} v=\sqrt[]{(4.5)^2+(9.8\times2.02)^2} \\ =20.3\text{ m/s} \end{gathered}[/tex]
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
[tex]R=u_x\times t[/tex]
Substitute the known values in the above exprssion as
[tex]\begin{gathered} R=(4.5)\times2.02 \\ =9.09\text{ m} \end{gathered}[/tex]
Hence, the distance of the ball lands from the tower is R = 9.09 m
