• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Projectile Motion question (1 Viewer)

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,256
Gender
Male
HSC
2012
Hey guys,

This question is from Surfing Physics and I've got no clue how to solve it...

A projectile fired up into the air from the top of a 75m high cliff hits the ground 500m out from the base.

Find the initial velocity, time of flight and maximum height.

If anyone can give a worked solution, it would be greatly appreciated :)
 

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
In this situation, you can not determine the initial velocity, time of flight, maximum height as a numeral, only in terms of variables
 

hscishard

Active Member
Joined
Aug 4, 2009
Messages
2,033
Location
study room...maybe
Gender
Male
HSC
2011
Prob was fired horizontally and not up into the air so max height was just the cliff. If so you can find the time of flight first, then you can determine the initial velocity by considering the range.

Is that the exact question but? if so, bleakarcher is right on the variables
 

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
it was fired up into the air lol. if it was horizontally fired:

Let the intial velocity be V and time of flight be T.
S(x)=u(x)t
When t=T, S(x)=500,
T=500/u(x)
S(y)=u(y)t+(1/2)a(y)t^2
S(y)=u(y)t-4.9t^2, where g=-9.8m/s^2
When t=T, S(y)=-75,
u(y)T-4.9T^2=-75
As T=500/u(x),
500u(y)/u(x)-4.9*500^2/u(x)^2=-75
Assuming the projectile is shot horizontally:
u(x)=Vcos0=V, u(y)=Vsin0=0
Hence:
4.9*500^2/V^2=75
V=sqrt[4.9*500^2/75]=127.8m/s
 

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,256
Gender
Male
HSC
2012
Thanks guys, I thought there was no way to work it out. The answer was velocity is 65 ms at 39.7 degrees so there must be something missing like time of flight. That was the question word for word BTW.

Cheers
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top