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A circus performer is shot out of a cannon at a 50° angle. He reaches a height of 25m. What was the initial velocity?

a. How long will he fly before he hits the safety net set at ground level with a hole underneath it. (The circus wanted it to look like he was going to hit the ground and then he bounces, for the thrill effect)

User Eldho
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1 Answer

5 votes

Answer:

28.9 m/s

4.52 s

Step-by-step explanation:

Given in the y direction:

s = 25 m

u = U sin 50°

v = 0 m/s

a = -9.8 m/s²

Find: U

v² = u² + 2as

0² = (U sin 50°)² + 2 (-9.8) (25)

U = 28.9 m/s

Given in the y direction:

s = 0 m

u = 28.9 sin 50° = 22.1 m/s

a = -9.8 m/s²

Find: t

s = ut + ½ at²

0 = (22.1) t + ½ (-9.8) t²

t = 4.52 s

User Corazon
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