Asked by James
A ball is thrown from a building standing 30m high, landing 10m away from the building 3.2 seconds after it was thrown. Find the angle above the horizontal and the initial velocity of the ball
Answers
Answered by
Damon
0 = 30 + Vi (3.2) - 4.9 (3.2)^2
Vi = 6.31 up
10 = u (3.2)
u = 3.13
tan theta = 6.31/3.13
theta = 63.6 degrees up from horizontal
speed = sqrt(3.13^2 + 6.31^2)
= 7.04
Vi = 6.31 up
10 = u (3.2)
u = 3.13
tan theta = 6.31/3.13
theta = 63.6 degrees up from horizontal
speed = sqrt(3.13^2 + 6.31^2)
= 7.04
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