# Homework Help: I'm trying to find max height of a projectile

1. Mar 29, 2004

### starlight1834

I'm trying to find max height of a projectile and I only know the angle of projection and the time it is in the air. No clue where to start working this problem...don't I need initial velocity to start?

2. Mar 29, 2004

### himanshu121

Welcome to PF. Here we will provide u with clues/hints if u show ur try.

At max height there will be no component of velocity in y direction so Vy=0
Eqn of trajectory:
$$y=xtan(\theta) - \frac{gx^2}{2V_0^2 cos^2(\theta )}$$

Diff above to get y_max
Or

$$y=v_0 sin(\theta) t- 0.5 gt^2$$
for $$y_{max} : t= \frac{v_0 sin\theta }{g} = 0.5 T$$

Where T is the time in Air
u wont need initial velocity
$$T= 2v_0 sin\theta * g^{-1}$$
so basically u know initial velocity as well

Last edited: Mar 29, 2004
3. Mar 29, 2004

### starlight1834

I still don't see how I don't need the initial velocity. I can plug everything else into the equation. I'm using the equation y=v*sin(angle)*t +.5*g*t
Is this the right equation? If so, why don't I need the velocity, and how do I get around it?

4. Mar 29, 2004

### starlight1834

negative max height?

I used the equations and found the initial velocity, and it was negative, so then when I put everything back into the y-max equation I got a negative number for the y-max. I don't think that's right, where did I go wrong? Please help ASAP b/c I have class in 45 min!

5. Mar 29, 2004

### himanshu121

there should be -g coz velocity is in opposite direction