A ball is thrown horizontally from the top of a building 109.53 m high. the ball strikes the ground at point 65 m horizontally away from and below the point of realease. What is the speed of the ball just before it strikes the ground

V = 48.4m/s, θ = 73.6° below the horizontal.

Explanation:

Given h = 109.53m, x = range = 65m,

θ = 0°

This problem involves the concepts of projectile motion.

Let yo = h = 109.53m

y = final position on the y axis. = 0m (ground level)

y = yo + vosinθ – 1/2gt²

0 = 109.53 + vosin0° – 1/2×9.8×t²

0 = 109.53 – 4.9t²

4.9t² = 109.53

t² = 109.53/4.9 = 22.35

t = √22.35 = 4.73s

So it takes the ball t = 4.73 seconds to get to the ground from the launch point.

x = (vocosθ)×t

65 = (vocos0°)×4.73

65/4.73 = vo

Vo = 13.7m/s

Vx = Vox = Vocosθ = 13.7cos0° = 13.7m/s

Vy = Voy – gt = Vosinθ – gt

Vy = 13.7sin0° – 9.8×4.73 = –46.4m/s

V = √(Vy² + Vx²) = √(-46.4² + 13.7²)

V = 48.4m/s

θ = Tan-¹(vy/vx) = tan-¹(-46.4/13.7) = -73.6°

θ = 73.6° below the horizontal

V = 48.4m/s

the answer is kevlar