pyh motion

2011-01-11 12:28 am

1,Given the acceleration due to gravity on Moon surface is 1.622 ms-2.
A stone is thrown upwards from the groud with an initial volocity of 20 ms-1
find the average velocity of the stone for the upward journey.

2. Given the acceleration due to gravity on Mercury surface is 3,7 ms-2 A stone is thrown upwards from the ground and reaches a height of 40 m
find the time needed for the stone back the ground.

3. Given the acceleration due to gravity on Mercury surface 3.7 ms-2 A stone released of 100m find the average veloctiy of the stone for the upward journey.

4. Given the acceleration due to gravity on Earth Surface is 10ms-2. A stone is thrown upwards from the ground and reaches a height of 40m
a) find the inital velocity of the stone
b) find the time needed fro the stone to reached the highest postion
c) find the velocity of the stone just before it hits the ground
d) find the time needed for the stone to back to the ground
e) find the average velocity of the stone for i) the upward jouney ii) the whole journey.
f) find the average acceleration of the stone for 1) the upward journey ii) the whole journey.

回答 (1)

天同

2011-01-11 1:03 am

✔ 最佳答案

1. Since the final velocity = 0 m/s
average velocity in the upward journey = 20/2 m/s = 10 m/s

2. Use equation of motion: s = vt - (1/2)at^2 to find the time taken for the stone to reach the higest point of 40m.
40 = (0).t - (1/2).(-3.7)t^2
solve for t gives t = 4.65 s
By symmetry, the time of fall from a height of 40m to ground = 4.65 s

3. Use equation of motion: v^2 = u^2 + 2a.s
0 = u^2 + 2(-3.7).(100)
u = 27.2 m/s
Average velocity = 27.2/2 m/s = 13.6 m/s

4.(a) use equation: v^2 = u^2 + 2a.s
0 = u^2 + 2(-10).(40)
u = 28.28 m/s
(b) Use v = u + at
0 = 28.28 + (-10).t
t = 2.828 s
(c) By symmetry, velocity just hit the ground = initial upward velocity = 28.28 m/s
(d) By symmetry, time to fall back to ground = time for the upward journey = 2.828 s
(e) (i) average velocity in upward journey = 28.28/2 m/s = 14.14 m/s
(ii) since for the whole upward and downward journey, displacement = 0 m, hence, average velocity = 0 m/s
(f) The acceleration is constant throughout the journey, which is 10 m/s2 in the downward direction.

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