Question

A golf ball is hit upward at an angle from the horizontal and speed v=40m|s.it reaches a maximum height of 10m as illustrated above.assume the ballistic trajectory starts from ground level and ignore air resistance..assume the shot is made on a wide level field...

What is the initial angle?

Answer 1

Explanation: v = 40 m/s, H = 10m, g = 9.8 m/s2

H = (v * sinθ)^2 / 2g

H = v^2 * (sinθ)^2 / 2g

H = (40)^2 * (sinθ)^2 / 2g

H = 1600 * (sinθ)^2 / 2g

Cross multiply

2gH = 1600 * (sinθ)^2

2 * 9.8 * 10 = 1600 * (sinθ)^2

196 = 1600 * (sinθ)^2

Make (sinθ)^2 the subject of the formula

(sinθ)^2 = 196 / 1600

Take the square root of both sides

√(sinθ)^2 = √( 196 / 1600)

sinθ = 0.35

θ = sin^-1(0.35)

θ = 20.48731511

Approx. 20.49

I hope this was helpful