A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landing is angled at 60 degree with the roof.

Question

a) find the horizontal distance d it travels.
b) what is the magnitude of the balls initial velocity?
c) what is the angle (relative to the horizontal) of the balls initial velocity?

Answer 1

In y to find initial v
20 = vi(4) +.5(-9.8)(4sqr)
so vi = 24.6
In y to find final v
vf = vi + at = 24.6 -(9.8*4) = -14.6
In x since angle = 60
vf = 14.6/tan60 = 8.4
In x vi=vf = 8.4
x = vi t = 8.4 * 4 = 33.7
Using pythagorean with initial x and y values
v = sqrt(8.4sqr + 24.6sqr) = 26.0
Initial angle
tan-1(24.6/8.4) = 71.1