An outfielder throws a baseball with an initial speed of 30 m/s at an angle of 13° to the horizontal. The ball leaves his hand from a height of 1.75 m. How long is the ball in the air before it hits the ground?

Answer 1

Upward motion:
v(y) = v(oy)- g•t1.
At the highest point v(y) = 0,
t1 =v(oy)/g = v(o) •sinα/g =
=30•sin 13º/9.8 = 0.69 s.
hₒ = v(oy) •t1- g•t1²/2= v(o)•sinα•t1 - g•t1²/2=
=30•sin 13º•0.69 – 9.8•(0.69)²/2 = 4.66-2.33 =2.33 m.
H = h + hₒ =1.75 +2.33 = 4.08 m.
v(x) = v(ox) = v(o) •cosα = 30•cos13º = 29.23 m/s.
Downward motion:
The time of downward motion is determined by the time of vertical motion from the height H:
H=g•t2²/2,
t2 =sqrt(2•H/g) =
=sqrt(2•4.08/9.8) =0.91 s.
t =t1+t2 = 0.69 +0.91 = 1.6 s.