You are Here: Home >< Maths

# M2- Projectiles Help watch

1. 17) A child throws a ball horizontally from a window 6m above horizontal ground. The ball just clears a vertical wall 2.5m high and 12m from the house. By modelling the ball as a particle calculate the speed of projection. The ball hits the ground at Q. Calculate the distance of Q from the wall.

Vertically:
s=12
a=0

Horizontally:
s=2.5
a=9.8
u=0

Resolving vertically: s=ut+1/2at^2
t=5/7s

Resolving horizontally: s=ut+1/2at^2
12=5/7u
u=16.8m/s

b) Horizontally:
v=16.8
u=16.8
a=0

Vertically:u=0
a=9.8
s=5

Resolving vertically: s=ut+1/2at^2
5=4.9t^2
t=1.01s

Resolving horizontally: s=ut+1/2at^2
=16.8*1.01
=16.97

Ergo, Horizontal distance from wall=16.97m-12=4.97

The textbook has the answer of 4.97m. However, what I am asking is that it asks for the distance of Q from wall, this is not the horizontal disance. So shouldn't it be:

Root of 4.97^2+2.5^2

18) A tennis ball is served horizontally at a speed of 24m/s from a height of 2.7m. The net is 1m high and 12m horizontally from the server. Model the ball as a particle and hence determine whether the ball cclears the net and if so by what distance.

How do you do 18? I have little idea.
2. First - Didn't read all the working, but it just means the horizontal distance. That's the shortest distance to the wall. You worked out the distance to the top of it I think?)

Second - So you can look at horizontal and vertical motion. You know horizontal distance and speed - can you see how to use this? And then how you can use what you find in the vertical motion?
Spoiler:
Show
find the time at which the ball has gone 12m horizontally
Spoiler:
Show
use this time and the acceleration of gravity to find how far it's fallen
3. (Original post by gcseeeman)
First - Didn't read all the working, but it just means the horizontal distance. That's the shortest distance to the wall. You worked out the distance to the top of it I think?)

Second - So you can look at horizontal and vertical motion. You know horizontal distance and speed - can you see how to use this? And then how you can use what you find in the vertical motion?
Spoiler:
Show
find the time at which the ball has gone 12m horizontally
Spoiler:
Show
use this time and the acceleration of gravity to find how far it's fallen
I did it as such:

Horizontal:
u=24
v=24
a=0
s=12
t=?

s=ut+1/2at^2
12=24t
t=2

Now what?
4. (Original post by Deep456)
I did it as such:

Horizontal:
u=24
v=24
a=0
s=12
t=?

s=ut+1/2at^2
12=24t
t=2

Now what?
If 12=24t, t isn't 2 (I guess you see the silly mistake). But use that for vertical motion and see how far it falls in that time.
5. (Original post by gcseeeman)
If 12=24t, t isn't 2 (I guess you see the silly mistake). But use that for vertical motion and see how far it falls in that time.
Ah yes, that was a school boy error.

### Related university courses

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: January 16, 2010
The home of Results and Clearing

### 1,287

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Keele University
Sun, 19 Aug '18
2. University of Melbourne
Sun, 19 Aug '18
3. Sheffield Hallam University
Tue, 21 Aug '18
Poll
Useful resources

### 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