Projectile Motion Calculation Question

[minifhncc]

Homework Statement

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

Homework Equations

$$\Delta$$x=u_xt

$$\Delta$$y=u_yt + 1/2 * a_yt²

u_y=usin$$\theta$$

The Attempt at a Solution

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

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

u_x=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

u_y=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 a lot if you can help me

Cheers

 

[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.

 

[minifhncc]

Hi,

Thanks for help but why do I need to make it -1/2 a_yt²?

Thanks