A baseball is thrown from the roof of a 22.6 m tall building with an initial vel

Question

A baseball is thrown from the roof of a 22.6 m tall building with an initial velocity of magnitude 10 m/s directed at an angle of 53.1° above the horizontal. At a later time the ball hits the ground. Assume there is no air resistance.

A) Choose all the energy types which change from the initial state (thrown from the roof) to the final state (just about to hit the ground).

Elastic Potential

Energy Kinetic Energy

Internal Energy

Gravitational Potential Energy

B) Use energy conservation to solve for the speed of the baseball just before it hits the ground.

Explanation / Answer

solution:

As the baseball is thrown from the roof of a building of height 22.6 m with an initial velocity of magnitude 10 m/s , the base ball possesses kinetic energy (KE) as well as potential energy (PE)

Initial KE=(1/2)mu^2 =(1/2)m*10*10=50 m J

Initial PT =mgh=m*9.8*22.6= 221.48 m J

Initial total energy E= KE + PE

Initial total energy E= 50 m + 221.48 m =271.48 m J

Initial total energy E= 271.48m J

When the ball strikes the ground, its final PE is zero and its final KE is (1/2)mV^2, where V is its velocity when it strikes the ground.

Final total energy E! = final KE + final PE

Final total energy E! =(1/2)mV^2 + zero

Ignoring air resistance, the total energy remains conserved

Final total energy E! = initial total energy E

(1/2)mV^2= 271.48 m J

cancelling 'm'

V^2 =2*271.48 =542.96

V = sq rt 542.96 =23.30 m/s

The speed of the ball just before it strikes the ground is 23.30 m/s ans

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