A baseball is thrown from the roof of 21.0 -tall building with an initial veloci

Question

A baseball is thrown from the roof of 21.0 -tall building with an initial velocity of magnitude 13.7 and directed at an angle of 50.4 above the horizontal.
A)
What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.
B)
What is the answer for part (A) if the initial velocity is at an angle of 50.4 below the horizontal?
C)
If the effects of air resistance are included, will part (A) or (B) give the higher speed?
The part (A) will give the higher speed.

The part (B) will give the higher speed.

Explanation / Answer

use conservation of energy

Total intial energy = total final energy

intially ball is at height h and have intial velocity u

finally ball is at zero height and have v velocity before striking the ground

so

1/2*m*u2+m*g*h=1/2m*v2+m*g*(0)

so we have

v = (u2+2gh)

a) u=13.7 and h=21m

v=((13.7)2+2*9.8*21) = 24.48m/s

b) the speed will not change with the ball direction

it only changes with magnitude so same 24.48m/s

c) effect of air resistance included external forces are acting on the ball so energy will be consumed in the process while coming down part of intial enrgy will only transfer to the ball at last moment and some will be lost doing work for resistance so velocity will drop in both cases

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