# Projectile motions help

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

How do I answer this question?

a) A rugby player is aiming for a conversion. He kicks the ball at 15ms−1 at an angle of 50∘ to the horizontal. At the time, he is 20m from the posts. How much time will the ball take to reach the posts?

b) How high will the ball be when it reaches the posts?

I've been struggling on this question for ages. So any answers and help would be really helpful...

a) A rugby player is aiming for a conversion. He kicks the ball at 15ms−1 at an angle of 50∘ to the horizontal. At the time, he is 20m from the posts. How much time will the ball take to reach the posts?

b) How high will the ball be when it reaches the posts?

I've been struggling on this question for ages. So any answers and help would be really helpful...

0

reply

Report

#2

(assuming no air resistance)

use trig to get the horizontal component of the initial velocity... this will get you the time to reach the posts v=s/t

use trig to get the vertical component of the initial velocity

then use suvat with acceleration = g to get the vertical displacement of the ball after the time you found in the earlier section

use trig to get the horizontal component of the initial velocity... this will get you the time to reach the posts v=s/t

use trig to get the vertical component of the initial velocity

then use suvat with acceleration = g to get the vertical displacement of the ball after the time you found in the earlier section

0

reply

(Original post by

(assuming no air resistance)

use trig to get the horizontal component of the initial velocity... this will get you the time to reach the posts v=s/t

use trig to get the vertical component of the initial velocity

then use suvat with acceleration = g to get the vertical displacement of the ball after the time you found in the earlier section

**Joinedup**)(assuming no air resistance)

use trig to get the horizontal component of the initial velocity... this will get you the time to reach the posts v=s/t

use trig to get the vertical component of the initial velocity

then use suvat with acceleration = g to get the vertical displacement of the ball after the time you found in the earlier section

Then moving onto part b, I calculated the the vertical component of V which is '11.49 m/s'.

Subbing all this into the suvat equation of 's = ut + at^2 and using the value of time from the earlier part along with '11.49 m/s' gives an answer of 2.8 to 1dp.

0

reply

**Joinedup**)

(assuming no air resistance)

use trig to get the horizontal component of the initial velocity... this will get you the time to reach the posts v=s/t

use trig to get the vertical component of the initial velocity

then use suvat with acceleration = g to get the vertical displacement of the ball after the time you found in the earlier section

A cricket batsman hits a ball at a speed of 27m/s at an angle of 60 degrees to the horizontal. How far would you have to stand in order to catch it (assuming you have to catch it before it hits the ground)?

How would I attempt this one since I don't know time or dist???

0

reply

Report

#5

(Original post by

One other Q I have is how would you do this too?

A cricket batsman hits a ball at a speed of 27m/s at an angle of 60 degrees to the horizontal. How far would you have to stand in order to catch it (assuming you have to catch it before it hits the ground)?

How would I attempt this one since I don't know time or dist???

**Yatayyat**)One other Q I have is how would you do this too?

A cricket batsman hits a ball at a speed of 27m/s at an angle of 60 degrees to the horizontal. How far would you have to stand in order to catch it (assuming you have to catch it before it hits the ground)?

How would I attempt this one since I don't know time or dist???

this time around you'd look at the vertical component of velocity first & find the the time(s) taken for the ball to reach 'catching height'. if you assume the batsman hits the ball at knee height and the fielder catches the ball at chest height then there might be a couple of solutions - one with the ball still on it's way up to the maximum height and one with the ball on it's way back down again.

the first case would correspond to the wicket-keeper making a quick reaction catch close to the batsman and the second case to a more conventional catch in wide field.

you'd use the time (s) and the horizontal component of velocity to find the distance horizontally.

0

reply

Report

#6

(Original post by

well you still need to find t (common theme to this type of question)

this time around you'd look at the vertical component of velocity first & find the the time(s) taken for the ball to reach 'catching height'. if you assume the batsman hits the ball at knee height and the fielder catches the ball at chest height then there might be a couple of solutions - one with the ball still on it's way up to the maximum height and one with the ball on it's way back down again.

the first case would correspond to the wicket-keeper making a quick reaction catch close to the batsman and the second case to a more conventional catch in wide field.

you'd use the time (s) and the horizontal component of velocity to find the distance horizontally.

**Joinedup**)well you still need to find t (common theme to this type of question)

this time around you'd look at the vertical component of velocity first & find the the time(s) taken for the ball to reach 'catching height'. if you assume the batsman hits the ball at knee height and the fielder catches the ball at chest height then there might be a couple of solutions - one with the ball still on it's way up to the maximum height and one with the ball on it's way back down again.

the first case would correspond to the wicket-keeper making a quick reaction catch close to the batsman and the second case to a more conventional catch in wide field.

you'd use the time (s) and the horizontal component of velocity to find the distance horizontally.

0

reply

Report

#7

(Original post by

i used 1.5m as my vertical height but i got answer 63.45m. isaac physics accepts 64m. bit ambiguous.

**lyds01**)i used 1.5m as my vertical height but i got answer 63.45m. isaac physics accepts 64m. bit ambiguous.

0

reply

Report

#8

(Original post by

Okay, so calculating the horizontal component of V gives '9.6418.. m/s' which then gives a a time of 2.074 s from '20/9.6418'

Then moving onto part b, I calculated the the vertical component of V which is '11.49 m/s'.

Subbing all this into the suvat equation of 's = ut + at^2 and using the value of time from the earlier part along with '11.49 m/s' gives an answer of 2.8 to 1dp.

**Yatayyat**)Okay, so calculating the horizontal component of V gives '9.6418.. m/s' which then gives a a time of 2.074 s from '20/9.6418'

Then moving onto part b, I calculated the the vertical component of V which is '11.49 m/s'.

Subbing all this into the suvat equation of 's = ut + at^2 and using the value of time from the earlier part along with '11.49 m/s' gives an answer of 2.8 to 1dp.

0

reply

Report

#9

Quick question i know im super late but i just got set this question for homework - why do you use 1.5m as the vertical height, where did you get that from, if you dont mind me asking?

0

reply

Report

#10

(Original post by

Quick question i know im super late but i just got set this question for homework - why do you use 1.5m as the vertical height, where did you get that from, if you dont mind me asking?

**Lina_Amiour**)Quick question i know im super late but i just got set this question for homework - why do you use 1.5m as the vertical height, where did you get that from, if you dont mind me asking?

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top