Answer:
θ ≈ 65°
Explanation:
From the question given above, the following data were obtained:
Maximum height (H) = 8 m
Range (R) = 15 m
Initial velocity (u) = v
Angle θ =?
H = u²Sine²θ / 2g
8 = v²Sine²θ / 2g
Cross multiply
8 × 2g = v²Sine²θ
16g = v²Sine²θ
Divide both side by Sine²θ
v² = 16g / Sine²θ.... (1)
R = u²Sine2θ / g
15 = v²Sine2θ / g
Cross multiply
15 × g = v²Sine2θ
15g = v²Sine2θ
Divide both side by Sine2θ
v² = 15g / Sine 2θ... (2)
Summary:
v² = 16g / Sine²θ.... (1)
v² = 15g / Sine 2θ... (2)
Equate equation 1 and 2
16g / Sine²θ = 15g / Sine 2θ
16 / Sine²θ = 15 / Sine 2θ
Recall:
Sine²θ = SineθSineθ
Sine 2θ = 2SineθCosθ
16 / Sine²θ = 15 / Sine 2θ
16 / SineθSineθ = 15 / 2SineθCosθ
16 / Sineθ = 15 / 2Cosθ
Cross multiply
15 × Sineθ = 16 × 2Cosθ
15Sineθ = 32Cosθ
Divide both side by Cosθ
15Sineθ / Cosθ = 32
Divide both side by 15
Sineθ / Cosθ = 32/15
Recall:
Sineθ / Cosθ = Tanθ
Sineθ / Cosθ = 32/15
Tanθ = 32/15
Tanθ = 2.1333
Take the inverse of Tan
θ = Tan¯¹ 2.1333
θ ≈ 65°
