You throw a stone upward from the top of a building at an angle of 25° to the horizontal and with an initial speed of 15 m/s. If the stone is in flight for 3.0 s, how tall is the building? How far from the base of the building does the stone.

11.025 m

87.75 m

Explanation:

Time of flight (T) of a projectile = 2U (sin∅) / g

Where U = initial Velocity, g = acceleration due to gravity, ∅ = angle of projection.

Make ∅ the subject of the the equation,

∅ = sin⁻¹[ (T * g) / 2U]

Where U = 15m/s, T = 3.0 s, g = 9.8 m/s²

∅ = sin⁻¹[ (3 * 9.8) / (2*15) ]

∅ = sin⁻¹ (29.4/30)

∅ = sin⁻¹ (0.98) = 78.52°

Using the formula for maximum height of a projectile

H = U²sin²∅/2g

H = 15² (sin²78.52) / 2 * 9.8

H = 225 (0.98 * 0.98) / 19.6

H = 216.09/19.6

H = 11.025 m

Range (R) = U²sin2∅/g

R = 15²sin (2*79.52) / 9.8

R = 225 (0.39)

R = 87.75 m

∵ the building is = 11.025 m tall and the base of the building is 87.75m away from where the stone landed.