A football is thrown at a angle into the air for 6s and lands 150m away at what was the angle released

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°

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