Footballer (soccer) Ronaldo fires a shot on goal from 18 m away. It strikes the crossbar (the
top of the goal) 2.4 m above the ground 0.7 s later having only a horizontal component of
velocity. What is the magnitude of the velocity at which the ball is kicked?
The answer should be 26.6 m/s (59.5 mph)
Physics Question?
2018-02-06 9:57 am
回答 (2)
2018-02-06 10:41 am
✔ 最佳答案
Horizontal V component = (18/0.7) = 25.7m/sec.Initial vertical V component = sqrt.(2gh) = sqrt.(19.6 x 2.4) = 6.86m/sec.
Initial V = sqrt.(6.86^2 + 25.7^2) = 26.6m/sec.
2018-02-06 10:24 am
Horizontal direction:
Horizontal component of initial velocity = (18/0.7) m/s
Vertical direction: (Take all upward quantities to be positive.)
Displacement, s = 2.4 m
Final velocity, v = 0 m/s
Time taken, t = 0.7 s
Initial velocity = ? m/s
s = (1/2)(u + v)t
2.4 = (1/2) × (u + 0) × 0.7
Vertical component of initial velocity, u = (4.8/0.7) m/s
Magnitude of initial velocity = √[(18/0.7)² + (4.8/0.7)²] m/s = 26.6 m/s
Horizontal component of initial velocity = (18/0.7) m/s
Vertical direction: (Take all upward quantities to be positive.)
Displacement, s = 2.4 m
Final velocity, v = 0 m/s
Time taken, t = 0.7 s
Initial velocity = ? m/s
s = (1/2)(u + v)t
2.4 = (1/2) × (u + 0) × 0.7
Vertical component of initial velocity, u = (4.8/0.7) m/s
Magnitude of initial velocity = √[(18/0.7)² + (4.8/0.7)²] m/s = 26.6 m/s
2018-02-06 10:30 am
Vx = 18/0.7=25.7m/s
0.7Vy - 4.9(.7)^2 = 2.4 ---->Vy=6.9
|V| = √[(Vx)^2+(Vy)^2] = 26.6m/s
0.7Vy - 4.9(.7)^2 = 2.4 ---->Vy=6.9
|V| = √[(Vx)^2+(Vy)^2] = 26.6m/s
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