PHY MOTION

Shirley

2011-01-10 11:06 pm

1. Given the acceleration due to gravity on Eath surface is 10 ms-2 . A stone is thrown upwards from the ground with an inital velocity of 20 ms-1
a) find the time needed for the stone to reach the highest postions.
b) find the time needed for the stone to be back to the ground
c) find the veloctiy of the stone just before it hits the ground

2> given the acceleration due to gravity on Eath durface is 10ms-2. A stone is released from a height of 100m
a) find the velocity of the stone just before it hits the ground
b) find the time needed for the stone to reach the ground

3) A stone is thrown upwards from the ground on Venus with a velocity of 20ms-1 and It reaches a height iof 22.548m
a.Prove the acceleration due to gravity on the Venus surface is equal to 8 .87ms-2
b) Find the time need for the stone to reach the highest position

回答 (1)

天同

2011-01-10 11:48 pm

✔ 最佳答案

1. (a) Use equation of motion:v = u + at
with u = 20 m/s, a = -g(=-10 m/s2), v = 0 m/s, t = ?
hence, 0 = 20 + (-10)t
i.e. t = 2 s
(b) Use equation: v^2 = u^2 + 2as to find the height reached
0 = 20^2 + 2(-10)s
i.e. s = 20 m
Use equation: s = ut + (1/2)at^2 to find the time of fall
-20 = (0).t + (1/2)(-10)t^2
i.e. t= 2 s
(c) Use equation: v = u + at
v = 0 + (-10).(2) m/s
i.e. v = -20 m/s

2.(a) Use equation: v^2 = u^2 + 2as
with u = 0 m/s, a = -g(=-10 m/s2), s = -100 m, v = ?
hence, v^2 = (2)(-10)(-100) (m2s)^2
v = -44.72 m/s
(b) Use v = u + a.t
-44.72 = (-10)t
i.e. t = 4.472 s

3. (a) Use equation: v^2 = u^2 + 2a.s
with v = 0 m/s, u = 20 m/s, s = 22.548 m, a = ?
hence, 0 = 20^2 + 2a.(22.548)
a = 8.87 m/s2
The acceleration due to gravity on Venus is 8.87 m/s2
(b) Use equation: v = u + at
0 = 20 + (-8.87)t
i.e. t = 2.25 s

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