Asked by sea lively

A ball within an initial velocity of 5.55 m/s is shot at an angle of 30 degree above the horizontal using a launcher. The launcher is 1.11 high.

a. How long is the ball in the air, before it hits the ground?
b. How far from the launcher does the ball land on the ground?

Answers

Answered by scott
a. using the free-fall equation ... 0 = -1/2 g t^2 + t sin(30º) + 1.11
... solve for t

b. distance = 5.5 cos(30º) * t ... use t from a.
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