How do I solve this question?

Question

A 0.15-kg ball is thrown so that it leaves your hand 2.0 m above the ground and has a speed of 25 m/s at an angle of 45 degrees upward from the horizontal. What is its speed just before it hits the ground? How high does the ball rise?

(assume there is no friction)

Answer 1

Use the formula

Vf^2 - Vo^2 = 2gs

where

Vf = speed of ball as it hits the ground

Vo = vertical component of the initial velocity = 25(sin 45)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

s = height of ball from ground

Substituting values,

Vf^2 - [25(sin 45)]^2 = 2(9.8)(2)

Vf^2 = 2(9.8)(2) + [25(sin 45)]^2

Vf^2 = 39.2 + 312.4

Vf^2 = 351.6

Vf = 18.75 m/sec.

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The maximum height that a projectile attains is given by the formula

Y = Vo^2(sin^2A)/2g + 2 (since it initially is 2 meters above the ground)

and substituting values,

Y = [25^2(sin 45)^2]/(2 * 9.8)  + 2

After performing the required arithmetical process,

Y = 15.94 + 2

Y = 17.94 meters (from the ground)