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

Respuesta :

Answer:

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

(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;

  • initial velocity of the object, u = 42 m/s
  • angle of projection, = 60°
  • time of motion of the stone, t = 5.5 s

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

[tex]h_y = v_0_yt - \frac{1}{2} gt^2\\\\h_y = (v_0\times sin(\theta)) t \ - \ \frac{1}{2} gt^2\\\\h_y = (42\times sin(60))\times 5.5 \ - \ (0.5\times 9.8\times 5.5^2)\\\\h_y = 200.05 - 148.23\\\\h_y = 51.82 \ m[/tex]

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

The vertical component of the final velocity

[tex]v_f_y = v_0_y - gt\\\\v_f_y = (v_0 \times sine (\theta)) - gt\\\\v_f_y = (42\times sin(60) - (9.8\times 5.5)\\\\v_f_y = -17.53 \ m/s[/tex]

The horizontal component of the final velocity

[tex]v_x_f = v_0\times cos(\theta)\\\\v_x_f = 42 \times cos(60)\\\\v_x_f = 21 \ m/s[/tex]

The resultant of the velocity at point A;

[tex]v_f = \sqrt{v_x_f^2 + v_y_f^2} \\\\v_f = \sqrt{(21)^2 + (-17.53)^2} \\\\v_f = 27.36 \ m/s[/tex]

(c) The maximum height reached by the projectile;

[tex]H = \frac{v_o^2 sin^2(\theta)}{2g} \\\\H = \frac{[42 \times sin(60)]^2}{2\times 9.8} \\\\H = 67.50 \ m[/tex]

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