Jump to content
Sign in to follow this  

Help With physics problem.

Recommended Posts

Guest SNJERIN

A projectile was launched perpendicular to a 30deg slope at 20ms^-1. Find the distance between the launching position and landing position. 

 

I tried it many time and keep getting 20√3m, but doesn't match the answers at the back of the book. 

Share this post


Link to post
Share on other sites

It might be the case that the book is wrong, because I also got the same answer as you do. Are there any other information given? because 20√3m is the answer only if we assume that the landing place is of the same height as the launching place.

Share this post


Link to post
Share on other sites
Guest SNJERIN

Actually that was my assumption too. However it doesn't make any sense if it was on ground level and has slope of 30. They haven't stated anything about the hight of which it was thown from.

Share this post


Link to post
Share on other sites

that's really weird then. I can't think of any other reasons why the answer is not 20√3m using the given information. I think we should blame for the vague wording of the question then :)

Share this post


Link to post
Share on other sites

 

May be a dumb question, but how did you get 20*sqrt(3)? Could anyone please explain? 

 

 

post-115475-0-92873000-1411933671.png

 

From the diagram, it's clear that the vertical velocity is 20sin(60), and the horizontal velocity is 20cos(60)

 

Now assume that the projectile lands at the same height as where it was launched, then the vertical velocity of the projectile when it lands is -20sin(60) (the minus sign represents the opposite direction of the velocity compared to when it was launched). Also let g = -10 m/s^2 stand for acceleration due to gravity.

 

Now we substitute all the information into the formula v = u + at, we get -20sin(60) = 20sin(60) -10t, which yields the total time taken to be t = 2√3

 

Now using the horizontal velocity and the formula s = vt, we get the distance to be s = 20cos(60) * 2√3 = 20√3 m. And that is the answer :)

  • Like 1

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.