Projectile Motion Homework Help

1. Oct 6, 2011

loopsnhoops

1. The problem statement, all variables and given/known data

Find the initial velocity and the maximum height

θ = 28˚
horizontal range/distance/s = 68 m
time = 6.30 s

2. Relevant equations

suvat equations not so sure which one though

3. The attempt at a solution

I think to find the initial velocity I use the equation:
s=ut+(1/2)at^2

I think acceleration horizontally is 0 so:

u(h) = s(h)/t

2. Oct 6, 2011

aftershock

There is no acceleration in the horizontal direction so you can simply use s=vt.

Now you can solve for the horizontal component of velocity and you have an angle so using trig you should now be able to solve for the hypotenuse of the initial velocity.

For the maximum height just remember to only focus on the vertical components and use your kinematics equations.

3. Oct 6, 2011

SammyS

Staff Emeritus
Use the last equation, u(h) = s(h)/t to find the horizontal component of the velocity. Since the horizontal component of the velocity is constant, what does this tell you about the horizontal component of the initial velocity?

4. Oct 6, 2011

loopsnhoops

It is the same as the u(h)?

5. Oct 6, 2011

loopsnhoops

Lol I don't know any of this...

6. Oct 6, 2011

SammyS

Staff Emeritus
Yes. The horizontal component of the velocity doesn't change, so that is the same as the horizontal component of the initial velocity.

From that use some trig to find the initial velocity and the vertical component of the initial velocity.