A projectile is fired at 28 m/s at an angle of 33 degrees with the horizontal from a point 75 m above the ground.
a) How long did it take for the projectile to hit the ground below?
b) How far did the projectile travel horizontally?
Physics Problem?
2020-09-06 9:59 am
回答 (2)
2020-09-06 1:54 pm
✔ 最佳答案
a)Take g = 9.8 m/s²
Consider the vertical motion (uniform acceleration motion).
Take all upward quantities to be positive, and downward to be negative.
h = (u sinθ) t + (1/2) g t²
-75 = (28 sin33°) t + (1/2) (-9.8) t²
4.9t² - (28 sin33°)t - 75 = 0
t = {(28 sin33°) + √[(-28 sin33°)² - 4(4.9)(-75)]} / [2 (4.9)] s
t = 5.77 s ≈ 5.8 s
b)
Range = (u cosθ) t = (28 cos33°) (5.77) m = 135 m
2020-09-06 11:10 am
a) Time to reach max height, t₁ = (v sinθ)/g = (28*sin33)/9.8 = 1.56 s
Height until 75m, h₁ = (v²sin²θ/2g) = (33²sin²33)/(2*9.8) = 16.48 m
Height from 75m to ground, h₂ = vsinθ t + ½gt²
Total height, H = h₁ + h₂ = 16.48 + 75 = 91.48 m
H = ½gt₂²
t₂ = √(2H/g)
t₂ = √(2*91.48/(9.8))
t₂ = 4.32 s
Total time to hit the ground, t = t₁+t₂ = 1.56 + 4.32 = 5.88 s
(b) Range = vcosθ * t = 28 cos33 * 5.88 = 138 m
Height until 75m, h₁ = (v²sin²θ/2g) = (33²sin²33)/(2*9.8) = 16.48 m
Height from 75m to ground, h₂ = vsinθ t + ½gt²
Total height, H = h₁ + h₂ = 16.48 + 75 = 91.48 m
H = ½gt₂²
t₂ = √(2H/g)
t₂ = √(2*91.48/(9.8))
t₂ = 4.32 s
Total time to hit the ground, t = t₁+t₂ = 1.56 + 4.32 = 5.88 s
(b) Range = vcosθ * t = 28 cos33 * 5.88 = 138 m
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