- #1

maobadi

- 22

- 0

## 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/s

^{2}

## 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..