Projectile Motion Calculation Question

AI Thread Summary
The discussion revolves around calculating various aspects of projectile motion, specifically a projectile with a time of flight of 7.5 seconds and a range of 1200 meters. The horizontal velocity was correctly calculated as 160 m/s. For the maximum height, the user attempted to derive the necessary equations but ended up with a different answer than the book, indicating a possible error in their calculations. The user also expressed confusion about the need to modify the equation for maximum height and questioned why the acceleration term should be negative. Clarification on these points is sought to ensure accurate calculations.
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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


\Deltax=uxt

\Deltay=uyt + 1/2 * ayt2

uy=usin\theta

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

\Deltay=(36.75*7.5)+(0.5*9.8*7.52)

\thetay=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
 
Physics news on Phys.org
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.
 
Hi,

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

Thanks
 
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