Question

After leaving the ground at the angle 53.1 degree with the horizontal, a ball has the minimum speed 22.2 m/s.
a/ Find the maximum speed of the ball.
b/ Find the time when the ball reaches the hight of 50 m. Justify your answer.

Answer 1

Given that,

Angle = 53.1°

Speed = 22.2 m/s

Distance = 50 m

(a). We need to calculate the maximum velocity

Using formula of range

$H=\dfrac{v^2\sin^2\theta}{2g}$

$v^2=\dfrac{H2g}{\sin^2\theta}$

Put the value into the formula

$v=\sqrt{\dfrac{50\times2\times9.8}{\sin^(53.1)}}$

$v=35.0\ m/s$

(b). We need to calculate the time when the ball reaches the height

Using formula of time of flight

$t=\dfrac{2v\sin\theta}{g}$

Put the value into the formula

$t=\dfrac{2\times35.0\sin53.1}{9.8}$

$t=5.7\ sec$

Hence, (a). The maximum velocity is 35.0 m/s.

(b). The time when the ball reaches the height is 5.7 sec.