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Mechanics Maths Help!
Q. A ball is released from rest at a point which is 10m above a wooden floor. Each time the ball strikes the floor, it rebounds with three-quarters of the speed with which it strikes the floor. Find the greatest height above the floor reached by the ball the first time it rebounds from the floor.
A. a=9.8
u=0
Now what do I do?
I keep getting the answer for s as 4.4m (displacement) and the book has it as 5.6m. As it is a rebound do I subtract the 4.4m from 10m?
A. a=9.8
u=0
Now what do I do?
I keep getting the answer for s as 4.4m (displacement) and the book has it as 5.6m. As it is a rebound do I subtract the 4.4m from 10m?
jenny28
Yh you have got 4.4 displacement from the 10m and the question asks for height above the floor, so your right just subtract it from the 10 
Only thing I am confused is other times with s you don't subtract it, do you do it in this case because it is a rebound?
The method I use is to break it down into seperate equations, I don't know which way you'd prefer but your information first gives :
U = 0
S = 10
A = 9.8
V = ?
V^2 = U^2 + 2AS
That will give you V = root 196 which is 14.
Now you use 14 as the initial velocity for where the ball bounces back up *0.75 as it's 3 quarters which gives 10.5.
Your new information is
U = 10.5
A = -9.8ms^-2
V = 0
S = ?
Now you can plug that again into V^2 = U^2 + 2AS and rearrange the equation which gives 5.6, the answer you were looking for
.
U = 0
S = 10
A = 9.8
V = ?
V^2 = U^2 + 2AS
That will give you V = root 196 which is 14.
Now you use 14 as the initial velocity for where the ball bounces back up *0.75 as it's 3 quarters which gives 10.5.
Your new information is
U = 10.5
A = -9.8ms^-2
V = 0
S = ?
Now you can plug that again into V^2 = U^2 + 2AS and rearrange the equation which gives 5.6, the answer you were looking for
.xSkyFire
The method I use is to break it down into seperate equations, I don't know which way you'd prefer but your information first gives :
U = 0
S = 10
A = 9.8
V = ?
V^2 = U^2 + 2AS
That will give you V = root 196 which is 14.
Now you use 14 as the initial velocity for where the ball bounces back up *0.75 as it's 3 quarters which gives 10.5.
Your new information is
U = 10.5
A = -9.8ms^-2
V = 0
S = ?
Now you can plug that again into V^2 = U^2 + 2AS and rearrange the equation which gives 5.6, the answer you were looking for .
U = 0
S = 10
A = 9.8
V = ?
V^2 = U^2 + 2AS
That will give you V = root 196 which is 14.
Now you use 14 as the initial velocity for where the ball bounces back up *0.75 as it's 3 quarters which gives 10.5.
Your new information is
U = 10.5
A = -9.8ms^-2
V = 0
S = ?
Now you can plug that again into V^2 = U^2 + 2AS and rearrange the equation which gives 5.6, the answer you were looking for
.So it is
New question:
A particle P is projected vertically upwards from a point O with a speed 12m/s. One second after P has been projected from O, another particle Q is projected vertically upwards from O with speed 20m/s. Find the time between the instant that P is projected from O and the instant when P and Q collide.
This question I genuinely have no ieda.
A particle P is projected vertically upwards from a point O with a speed 12m/s. One second after P has been projected from O, another particle Q is projected vertically upwards from O with speed 20m/s. Find the time between the instant that P is projected from O and the instant when P and Q collide.
This question I genuinely have no ieda.
Anyone, struggling so much!
Deep456
Anyone, struggling so much!
Can you do the equation of motion for P, giving s in terms of the other variables?
Original post by xSkyFire
The method I use is to break it down into seperate equations, I don't know which way you'd prefer but your information first gives :
U = 0
S = 10
A = 9.8
V = ?
V^2 = U^2 + 2AS
That will give you V = root 196 which is 14.
Now you use 14 as the initial velocity for where the ball bounces back up *0.75 as it's 3 quarters which gives 10.5.
Your new information is
U = 10.5
A = -9.8ms^-2
V = 0
S = ?
Now you can plug that again into V^2 = U^2 + 2AS and rearrange the equation which gives 5.6, the answer you were looking for .
U = 0
S = 10
A = 9.8
V = ?
V^2 = U^2 + 2AS
That will give you V = root 196 which is 14.
Now you use 14 as the initial velocity for where the ball bounces back up *0.75 as it's 3 quarters which gives 10.5.
Your new information is
U = 10.5
A = -9.8ms^-2
V = 0
S = ?
Now you can plug that again into V^2 = U^2 + 2AS and rearrange the equation which gives 5.6, the answer you were looking for
.This was really helpful!!!
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- Mechanical Engineering
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- What should I do to improve my grades in maths?
- Imperial Msc Soil Mechanics
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- how do you revise for maths mechanics at as level?
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