# Projectile motion cannonball problem

1. Jan 28, 2009

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

A cannonball is fired with an angle of elevation of 70o. Its range is 1km. Neglect the air resistance and calculate:
1. the time the cannonball is in the air.
2. its initial velocity.
3. its range and height at 18s after firing
4. the other range at which the height was the same as calculated in 3 above.

2. Relevant equations

u = the initial velocity
ux = ucos70o
uy = u sin70o
a = -9.81m/s2

3. The attempt at a solution

My attempt.

Range = ucos70o x t
therefore
t = Range/ucos70o

At full range for the Y component.
t = (Vy - u sin70o)/-9.81
t = u sin70o)/9.81

Since the time are same.

Range/ucos70o = u sin70o/-9.81 ( Is this right???)

I am confused. Can someone give me hints...thanks in advance..

2. Jan 28, 2009

### Gib Z

You have an error for your analysis of the y component. You haven't justified by V_y is zero at full range, and in fact it isn't! Think about it.

To continue, you should use $$s = ut + \frac{1}{2}at^2$$.

Or if you remember the formula $$R = \frac{u^2 \sin 2\theta}{g}$$ that could be very useful.