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?

Reply 1

jenny28

14

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 :biggrin:

Reply 2

Newcastle456

OP

14

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 :biggrin:

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?

Reply 3

xSkyFire

19

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 :p:

.

Reply 4

Newcastle456

OP

14

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 :p:

.

So it is

Reply 5

Newcastle456

OP

14

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.

Reply 6

Newcastle456

OP

14

Anyone, struggling so much!

Reply 7

ghostwalker

17

Deep456

Anyone, struggling so much!

Can you do the equation of motion for P, giving s in terms of the other variables?

Reply 8

learner_101

5

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 :p:

.

This was really helpful!!!

Mechanics Maths Help!

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