Question

Q # 2: A horseshoe player releases the horseshoe at A with an initial speed Vo = 36 ft/sec. Determine the range for the launch angle ɵ for which the shoe will strike the 14-in. vertical stake.​

Answer 1

Given info : a horseshoe player releases the horseshoe at a speed , v₀ = 36ft/sec .

To find : the range for the launch angle θ for which the shoe will strike the 14 inch vertical stake.

solution : vertical displacement, H = 14 inch = 1.167 ft

we know, H = u²sin²θ/2g

⇒ 1.167 ft = (36 ft/sec)²sin²θ/2(32 ft/sec²)

[ acceleration due to gravity in ft/s² is 32ft/s²]

⇒ sin²θ = (64 × 1.167)/(36²) = 0.0576

⇒ sinθ = 0.24

after solving you get, sin2θ = 0.465

now the range = u²sin2θ/g = (36)²(0.465)/(2 × 32) = 9.41625 ft

therefore the range for which the shoe will strike the 14 inch vertical stake is 9.41625 ft