A baseball is thrown from the roof of 21.1 m-tall building with an initial veloc

Question

A baseball is thrown from the roof of 21.1 m-tall building with an initial velocity of magnitude 11.6 m/s and directed at an angle of 53.9 above the horizontal. Use = 9.80 m/s^2 .

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 53.9 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

a)1/2 m*11.6^2 +mg*21.1 =1/2 mv^2 hence v =23.41 m/s b)velocity will be same as part one as no angle is use in energy conservation c) part B will give higher speed as time of projectile is less in part b than part a hence considering constant air resistance for more time decrease in velocity will be more..

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