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Question A rocket is launched from the top of a tower that is a height of 30 metres above ground level.
The initial speed at launch is 20m/s at an angle of 30degrees above the horizontal ground level
Calculate 1. The time of flight to the ground
2. the speed of impact when it hits the ground
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please help me with this physics/mechanics question!! urgent
2007-03-07 2:30 pm
回答 (1)
2007-03-07 6:45 pm
✔ 最佳答案
Assume g = 10m/s^2(1)
By s = ut + (1/2)at^2,
-30 = (20sin30)t + (1/2)(-10)t^2
-30 = 10t - 5t^2
t^2 - 2t -6 = 0
(t - 1)^2 = 7
t = 1 + sqroot(7) = 1.837s
Therefore, time of flight = 1.837s
(2)
At time t = 1.837s,
vertical velocity is given by
v = u + at
v = 20sin30 - (10)(1.837)
v = 10 - 18.37
v = -8.37m/s
horizontal velocity remains unchanged at 20cos30 = 17.32 m/s
Therefore, speed of impact
= sqroot[(-8.37)^2 + 17.32^2]
= sqroot(70+300)
= 19.235m/s
收錄日期: 2021-04-26 13:23:33
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