✔ 最佳答案
Refer to the figure below.
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A)
Take g = 9.81 m/s²
Consider the horizontal motion (uniform velocity):
s(x) = 17.0 m
u(x) = u cos55.0° m/s
s(x) = u(x) t
17.0 = (u cos55.0°) t
t = 17.0 / (u cos55.0°) …… [1]
Consider the vertical motion (uniform acceleration):
s(y) = 8.0 m
u(y) = u sin55.0° m/s
a(y) = -9.81 m/s²
s(y) = u(y) t + (1/2) a t²
8.0 = (u sin55.0°) t + (1/2) (-9.81) t²
4.905 t² - (u sin55.0°) t + 8.0 = 0 …… [2]
Substitute [1] into [2]:
4.905 [17.0 / (u cos55.0°)]² - (u sin55.0°) (17.0 / u cos55.0°) + 8.0 = 0
(4309/u²) - 16.28 = 0
u = √(4309 / 16.28)
Initial velocity, u = 16.27 m/s ≈ 16.3 m/s (to 3 sig. fig.)
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B)
Consider the horizontal motion (uniform velocity):
v(x) = u(x)
v(x) = 16.27 cos55.0°
v(x) = 9.332 m/s
Consider the vertical motion (uniform acceleration):
v(y)² = u(y)² + 2 a(y) s(y)
v(y)² = (16.27 sin55.0°)² + 2 (-9.81) (8)
v(y)² = 20.67 m²/s²
v = √[v(x)² + v(y)²]
v = √(9.332² + 20.67]
Final velocity, v = 10.4 m/s
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C)
v(y) = -√v(y)²
v(y) = -√20.67
v(y) = -4.546 m/s
tanθ = v(y) / v(x)
tanθ = -4.546/9.332
θ = tan⁻¹(-4.546/9.332)
θ = -26.0°
Velocity direction of the rock just before hitting the wall = 26.0° below the horizontal.
