28. a stone is projected at a cliff of height h with an initial speed of 42.0 mls directed at angle θ0 = 60.0° above the horizontal. the stone strikes at a, 5.50 s after launching. find (a) the height h of the cliff, (b) the speed of the stone just before impact at a, and (c) the maximum height h reached above the grou

(a) The height of the cliff is 51.82 m

(b) The final velocity of the stone before it strikes A is 27.36 m/s

(c) The maximum height reached by the stone is 67.50 m

The given parameters;

(a) The height of the cliff is calculated as;

(b) The final velocity of the stone before it strikes A is calculated as;

The vertical component of the final velocity

The horizontal component of the final velocity

The resultant of the velocity at point A;

(c) The maximum height reached by the projectile;

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a) 51.8 m, b) 27.4 m/s, c) 142 m

Explanation:

Given:

v₀ = 42.0 m/s

θ = 60.0°

t = 5.50 s

Find:

h, v, and H

a) y = y₀ + v₀ᵧ t + ½ gt²

0 = h + (42.0 sin 60.0) (5.50) + ½ (-9.8) (5.50)²

h = 51.8 m

b) vᵧ = gt + v₀ᵧ

vᵧ = (-9.8)(5.5) + (42.0 sin 60.0)

vᵧ = -17.5 m/s

vₓ = 42.0 cos 60.0

vₓ = 21.0 m/s

v² = vₓ² + vᵧ²

v = 27.4 m/s

c) vᵧ² = v₀ᵧ² + 2g(y - y₀)

0² = (42.0)² + 2(-9.8)(H - 51.8)

H = 142 m