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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