Question

A baseball is hit with a speed of 29.0 m/s at an angle of 43.0 ∘ . It lands on the flat roof of a 14.0 m -tall nearby building.

If the ball was hit when it was 1.3 m above the ground, what horizontal distance does it travel before it lands on the building?

Express your answer using two significant figures.

Answer 1

In vertical direction:
d = (14.0 - 1.3) = 12.7 m
a = -9.8 m/s^2
vi=29*sin 43 = 19.8 m/s
use:
d = vi*t + 0.5*a*t^2
12.7 = 14.5*t + 0.5*(-9.8)*t^2
12.7 = 19.8*t - 4.9*t^2
4.9*t^2 - 19.8*t + 12.7 =
t= 3.24 s or t = 0.8 s
0.8 s corresponds to time going up and 3.24 s corresponds to time coming up
Ball will land on roof while falling down

In horizontal direction,
d = ?
t = 3.24 s
vx = 29* cos 43 = 21.2 m/s
d = vx*t
= 21.2 * 3.24
= 68.7 m
Answer: 68.7 m