Answer:
50°
Explanation:
From the question given above, the following data were:
Time of flight (T) = 6 s
Range (R) = 150 m
Acceleration due to gravity (g) = 9.8 m/s²
Angle of projection (θ) =?
Thus, the angle of projection (θ) can be obtained as follow:
T = 2uSineθ /g
6 = 2 × u × Sine θ/9.8
6 = u × Sine θ/4.9
Cross multiply
6 × 4.9 = u × Sine θ
29.4 = u × Sine θ
Divide both side by Sine θ
u = 29.4 / Sine θ
R = u² Sine 2θ/ g
150 = u² Sine 2θ/ 9.8
u = 29.4 / Sine θ
150 = (29.4 / Sine θ)² Sine 2θ /9.8
150 = 864.36 /Sine²θ × Sine 2θ /9.8
150 = 88. 2 × Sine 2θ / Sine²θ
Recall:
Sine 2θ = 2SineθCosθ
150 = 88. 2 × 2SineθCosθ / Sine²θ
150 = 88. 2 × 2Cosθ / Sineθ
150 = 176.4 × Cosθ / Sineθ
Cross multiply
150 × Sineθ = 176.4 × Cosθ
Divide both side by Cosθ
150 × Sineθ/Cosθ = 176.4
Divide both side by 150
Sineθ/Cosθ = 176.4/150
Sineθ/Cosθ = 1.176
Recall:
Sineθ/Cosθ = Tan θ
Tan θ = 1.176
Take the inverse of Tan
θ = Tan¯¹ (1.176)
θ = 49.6° ≈ 50°
Thus, the angle of projection is 50°
