A baseball is thrown from the roof of a 27.5 m tall building with an initial velocity of magnitude 16.0m/s and directed at an angle of 37 above the horizontal Using energy methods and ignoring air resistance, calculate the speed of the ball just before it strikes the ground.

Given Information:  

distance = y = 27.5 m  

Angle = θ = 37°

initial speed of the ball = v1 = 16 m/s  

Required Information:  

final speed of the ball = v2 = ?

v2 = 28.19 m/s

Explanation:  

We will write an equation in terms work done by the forces acting on the ball. We know that one force is gravitational force

W = mgy1 - mgy2

kinetic energy is given by

KE = 0.5mv1² so

0.5mv1² + mgy1 = 0.5mv2² + mgy2

Mass cancels out and y2 is zero so equation becomes

0.5v1² + gy1 = 0.5v2²

0.5v2² = 0.5v1² + gy1

v2² = 2*(0.5v1² + gy1)

v2 = √2*(0.5v1² + gy1)

Substitute the values

v2 = √2*(0.5(16)² + 9.8*27.5)

v2 = 28.19 m/s

Therefore, the speed of the ball is 28.19 m/s just before it strikes the ground.

where is the circuit diagram ?

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