Question

A ball leaves the ground at the speed of 37 m/s and at the angle 53.1 .
a/ Find the maximun and the minimum speed of the ball after leaving the ground. Explain.
b/ Find the time when the ball reaches the hight of 50 m. Justify your answer.

Answer 1

Given:

Initial velocity of ball leaving ground = 37 meters per second at an angle of 53.1 with ground.

To Find:

1)The Maximum and the Minimum speed of the ball after leaving the ground

2)The time when the ball reaches the height of 50 m.

Solution:

The ball leaves the ground at the speed of 37 meters per second at an angle of 53.1 follows a projectile motion.

We divide the motion of the ball in two directions : vertical direction and along the ground surface.

Resolve the speed of the ball .

$u_{1}$ - initial velocity in x direction.

$v_{1}$ - initial velocity in y direction.

$u_{1}= 37 Cos53.1 = 22 \frac{m}{s} \\\\v_{1} = 37 Sin53.1 = 30 \frac{m}{s}$

Maximum speed of ball is when it is just released from the ground because as the ball moves away from the ground, the velocity in vertical direction decreases due to the retardation caused by acceleration due to gravity.

Hence Maximum speed of ball = 37 meters per second.

Minimum speed of ball: Due to retardation in vertically downward direction , velocity in vertical direction goes on decreasing until it becomes 0 and the ball start returning back to the ground due to acceleration due to gravity.

There is no acceleration in horizontal direction , hence velocity in horizontal direction remains same .

Hence the minimum velocity of ball is equal to 22 meters per second.

2)Height = 50 m.

initial velocity in vertical direction = 30 meters per second.

using the equation of motion , calculate the velocity at height of 50 m.

$v^{2} - u^{2} = 2as$

Maximum height reached by ball:

v = 0 u = 30 s = H a = g ;

$30^{2} = 2(10)(H)\\ H= 45 m.$

The ball will never reach the height of 50 meters since the maximum hieght of the ball is only 45 meters.