Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
 You are Here: Home >< Maths

# M1 Question on kinematics watch

1. 1) A ball is projected vertically upwards from a point O with speek 14ms-1. Find the greatest height above reached by the ball?

S =?
u = 14
v = ?
a= -9.8
t =?

I don't think i have enough information to use suvat. In the mark scheme they assume that V=0 m/s but i do not understand why?

Could somebody please explain why they have assumed the final velocity to be 0.

2) A particle P is projected vertically upwards from a point X. Five seconds later P is moving downwards with 10ms-1. Find

a) the speed of the projection of p
b) the greatest height above X attained by P during its motion.

so the speed of the projection is 39m/s. So i then wrote out suvat

s=?
u= -39
v
a=-9.8
t

Here again they have assumed v to 0 but why?
2. (Original post by Appazap)
1) A ball is projected vertically upwards from a point O with speek 14ms-1. Find the greatest height above reached by the ball?

S =?
u = 14
v = ?
a= -9.8
t =?

I don't think i have enough information to use suvat. In the mark scheme they assume that V=0 m/s but i do not understand why?

Could somebody please explain why they have assumed the final velocity to be 0.
Because when you throw something like a ball upwards, it will go up and up until it reaches it's max point and then it goes back down. Precisely at that max point, the velocity must be 0 since the ball is stationary at that instant.

2) A particle P is projected vertically upwards from a point X. Five seconds later P is moving downwards with 10ms-1. Find

a) the speed of the projection of p
b) the greatest height above X attained by P during its motion.

so the speed of the projection is 39m/s. So i then wrote out suvat

s=?
u= -39
v
a=-9.8
t

Here again they have assumed v to 0 but why?
It would be +39 for , not negative.
The reason is the same as before.
3. (Original post by RDKGames)
Because when you throw something like a ball upwards, it will go up and up until it reaches it's max point and then it goes back down. Precisely at that max point, the velocity must be 0 since the ball is stationary at that instant.

It would be +39 for , not negative.
The reason is the same as before.

Okay this makes a lot of sense, thank you x

Reply
Submit reply
Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 9, 2018
Today on TSR

### How do I turn down a guy in a club?

What should I do?

Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.