Question

A baseball leaves the bat at 30.0m/s at 36.9 degrees (see above). at what two times is the ball at a height of 10.0m above the point it left the bat?

Answer 1

Range = (u^2 * sin of angle) ÷ g

acceleration due to gravity (g) = 9.8m/s
...

R = (u^2 sin 36.9) ÷ 9.8 = (900 * 0.600420225) ÷ 9.8

R = 540 ÷ 9.8 = 55.102 m ...

t = 55 ÷ 30 = 1.833s ... total flight time...

takeoff time and landing time...

tan = opp ÷ adj

tan 36.9 = 10 ÷ adjacent... = 0.750821238 = 10 ÷ a

a = 10 ÷ 0.8 = (10 * 10) ÷ 8 = 12.5m

55.102m = 1.833
12.5m = (12.5 ÷ 55.102) * 1.833 = 0.4158s

1st time = 0.42s ...at takeoff

2nd time = 1.833 - 0.42 = 1.413s ...at landing

Ans = 0.42s ...and 1.41s ...