Question

A baseball is thrown from the roof of 23.5 m -tall building with an initial velocity of magnitude 10.5 m/s and directed at an angle of 52.4° above the horizontal.

What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.

What is the answer if the initial velocity is at an angle of 52.4° below the horizontal?

If the effects of air resistance are included, will part (a) or (b) give the higher speed? Why?

Answer 1

A baseball is thrown from the roof of 23.5 m -tall building with an initial velocity of magnitude 10.5 m/s and directed at an angle of 52.4° above the horizontal.

a)

Let m be the mass of the baseball.

The potential energy of the baseball at the roof is
Ep = m×g×h = m×9.8×23.5 = m×230.3

The kinetic energy of the baseball at the roof is
Ek = m×v²/2 = m×10.5²/2 = m×55.1

The total mechanical energy of the baseball at the roof is
Em = Ep + Ek = m×230.3 + m×55.1 = 285.4×m

When the baseball strikes the ground, all energy is converted to kinetic energy, so
Ek = m×v²/2 = m×285.4
v²/2 = 285.4
v = 23.9 m/s < - - - - - - - answer (a)

b)

I didn't use the launching angle to solve part a), which means that it doesn't affect the result, so
v = 23.9 m/s < - - - - - - - answer (b)

c)

In (b), the ball has a shorter path through the air, so it will be slowed down less than the ball in part (a), so (b) has the higher speed. < - - - - - - - answer (c)