15. A ball was thrown from the ground at an angle 0 with a velocity v. The ball reached a maximum height of 8 m and a maximum range of 15 m. Calculate the value of 0. Let g=1 m/s2​

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°

answer: good job it's all correct !

explanation: