Projectile motion cannonball problem

[maobadi]

Homework Statement

A cannonball is fired with an angle of elevation of 70^o. 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.

Homework Equations

u = the initial velocity
u_x = ucos70^o
u_y = u sin70^o
a = -9.81m/s²

The Attempt at a Solution

My attempt.

Range = ucos70^o x t
therefore
t = Range/ucos70^o

At full range for the Y component.
t = (V_y - u sin70^o)/-9.81
t = u sin70^o)/9.81

Since the time are same.

Range/ucos70^o = u sin70^o/-9.81 ( Is this right?)

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

 

[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.