# Projectile Motion Calculation Question

1. Sep 29, 2008

### minifhncc

1. The problem statement, all variables and given/known data
A projectile has a time of flight of 7.5s and a range of 1200m. Calculate:
a) Its horizontal velocity (This one was done correctly, just posting for follow through reference)

b) Its maximum height

c) The velocity with which it is projected

2. Relevant equations
$$\Delta$$x=uxt

$$\Delta$$y=uyt + 1/2 * ayt2

uy=usin$$\theta$$

3. The attempt at a solution

a)
1200=ux*7.5
ux=160ms-1 (This is correct)

b)
7.5 = (2usin$$\theta$$)/9.8
2usin$$\theta$$=73.5 ---- Equation 1

ux=ucos$$\theta$$
160=ucos$$\theta$$
u = 160/cos$$\theta$$ ---- Equation 2

Sub Equation 2 into 1:
2(160/cos$$\theta$$)*sin$$\theta$$=73.5
320tan$$\theta$$=73.5
tan$$\theta$$=73.5/320
$$\theta$$=12degree 56 min (nearest min)

u=164.17ms-1

uy=usin$$\theta$$
=164.17*sin12deg56min
=36.75ms-1

$$\Delta$$y=(36.75*7.5)+(0.5*9.8*7.52)

$$\theta$$y=551.25m (This is my final answer but it isn't the answer in the book :()

c) Yeah u=164.17ms-1 But I can't get direction. I stated that direction was 12deg56min from ground, but it's wrong...

Thanks alot if you can help me

Cheers

2. Sep 29, 2008

### rl.bhat

To calculate maximum height , you have to modify the equation LaTeX Code: \\Delta y=uyt + 1/2 * ayt2 by LaTeX Code: \\Delta y=uyt - 1/2 * ayt2. And for time you have to take half the time of flight.

3. Sep 29, 2008

### minifhncc

Hi,

Thanks for help but why do I need to make it -1/2 ayt2?

Thanks