A ball is thrown from the top of a tower of height 63.8 m with a velocity of 30 m/s at an angle of 37° above the horizontal. The angle formed by the velocity of the stone with the horizontal when it hits the ground is (take g = 10 m/s²)

A ball is thrown from the top of a tower of height 63.8 m with a velocity of 30 m/s at an angle of 37° above the horizontal.
Given : g = 10 m/s², h = 63.8 m, u = 30 m/s, θ = 37°
Formula: for vertical motion
h = u sinθt - (1/2)gt²
Sol: Time taken to reach the ground,
h = u sinθt - (1/2)gt²
=> 63.8 = (30 sin 37°)t - (1/2) × 10 × t²
=> 63.8 = 18t - 5t²
=> 5t² - 18t + 63.8 = 0
=> t = 5.8 sec
X-component:
Initial velocity, u = u cosθ = u cos 37° = 30 × 0.80 = 24 m/s
Velocity after t sec, v = u + at = 2u + (0 × 5.8) (≈ acceleration along horizontal direction) = 24 m/s
Y-component
Initial velocity, u' = u sinθ = u sin 37° = 30 × 0.60 = 18 m/s
velocity after t sec, v' = u' + a't = 18 + (-10)(5.8) = -40 m/s
So, velocity after t sec in vector form, v = 24î - 40ĵ
Angle formed by the velocity of the stone with the horizontal
θ' = tan⁻¹(Vy/Vx) = tan⁻¹(40/24)
θ' = tan⁻¹(5/3) ≈ tan⁻¹(1.667)
θ' ≈ 59°
The closest option is tan⁻¹(53) which is approximately 59°.
Therefore, Option C is approximately correct. Note that there might be slight discrepancies due to rounding errors in the calculations and the provided options not being precise angles.