A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is made with an initial speed of 19.0m/s at an angle of 33.0 ∘ above the horizontal. How long is the ball in the air?

Answer 1

Use the following formula:

Vf=Vi^+gt

Where

t=?
Vi=(19m/s*Sin33º)
Vf=0
g=-9.8m/s^2

Solve for t,

Vi=(19m/s*Sin33º)+(-9.8m/s^2)t

0=10.35m/s-9.8m/s^2*t

-10.35m/s=-9.8m/s^2*t

(-10.35m/s)/9.8m/s^2=-9.8m/s^2/-9.8m/s^2*t

t=1.06 for the ball to reach the top of its trajectory.

2*t= total time=2*1.06=2.12s