if a ball is thrown up from the ground and takes 2.30 seconds before it hits the ground. A) what was its initial velocity b) what was its max velocity

A) The initial velocity is 11.27 m/s

B) The maximum velocity is 11.27 m/s

Explanation:

A) The question is with regards to kinematic motion under gravity

The the time it takes the ball to travel up and return back to the ground = 2.30 seconds

Therefore, from the kinematic equation of motion of the ball, under gravity, v = u - g·t, we have;

t = (v - u)/g

Where;

t = The total time of the motion of the ball = 2.30 seconds

v = The final velocity of the ball = 0 m/as at the maximum height

u = The initial velocity of the ball

g = The acceleration due to gravity = 9.8 m/s²

Therefore, given that we have;

The time it takes the ball to ascend = The time it takes the ball to descend

The time it takes the ball to ascend to maximum height = 2.30 second/2 = 1.15 seconds

Substituting the parameter values of the motion to maximum height, we get;

1.15 = (0 - u)/(-9.8)

-9.8 × 1.15 = -u

11.27 = u

The initial velocity, u = 11.27 m/s

B) Given that the ball experiences a deceleration on the way up, and that the motion of a parabola is symmetrical about the vertex, which is the maximum height, where the velocity is zero, we have that the initial velocity is equal to the final velocity which are both equal to the maximum velocity

Therefore, the initial velocity = The maximum velocity = 11.27 m/s.

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