A ball is shot from the top of a building with an initial velocity of 18...
A ball is shot from the top of a building with an initial velocity of 18 m/s at an angle θ = 47 ∘ above the horizontal.What are the x and y components of the initial velocity?If a nearby building is the same height and 55 m away, how far below the top of the building will the ball strike the nearby building?Express your answer to two significant figures and include the appropriate units.
Homework Answers
a)
vx = 18 cos 47 = 12.276 m/s
vy = 18 sin 47 = 13.164 m/s
======
b)
time taken by ball to reach the building
t = 55 / vx = 4.48 s
vertical displacement
h = - vy t + 0.5* gt^2
h = - 13.164* 4.48 + 4.9* 4.48^2
h = 39.37 m
=======
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