Gr 11, Projectile Motions question?

[lookingtolrn]

1. A soccer ball is kicked at 25m/s with a 25° angle to the ground. What is
a) time it takes the ball to reach max height
b) the max height of the ball
c) time it takes to land on the ground again
d) range the ball travels
e) final velocity of the ball (with angle)2.
v2= v2 +aΔt
Δd= .5(v2 + v1)Δt
Δd= v1Δt + .5aΔt²
Δd= v2Δt - .5aΔt²
v2²=v1² + 2aΔd

3. any of them would be helpful!

 

[PeterO]

1. A soccer ball is kicked at 25m/s with a 25° angle to the ground. What is
a) time it takes the ball to reach max height
b) the max height of the ball
c) time it takes to land on the ground again
d) range the ball travels
e) final velocity of the ball (with angle)


2.
v2= v2 +aΔt
Δd= .5(v2 + v1)Δt
Δd= v1Δt + .5aΔt²
Δd= v2Δt - .5aΔt²
v2²=v1² + 2aΔd




3. any of them would be helpful!

You could play around with this:
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/enapplet.html
And then check this
http://en.wikipedia.org/wiki/Projectile_motion
Then work through this:
http://tutor4physics.com/projectilemotion.htm
and then you should have it.

 

[McKeavey]

a) Hard to explain over this.. but
First you need to break the velocity into x and y components.
So..

Vy = 25m/sSin25 = 10.56m/s
Vx = 25m/sCos25 = 22.65m/s

K now so when you draw the projectile motion out, the ground is level all the way throughout the projectile motion.
So maximum point of this motion would be the highest peak.
This also means that we are only looking at half the time of the entire motion.
So from when it was kicked (Vi), to when it is at its highest point (Vf), we can use the equation Vfy^2 = Viy^2 = 2ad
Vf = 0m/s (Because Velocity is 0m/s when it is at the highest peak [Not moving])
Viy = 10.56m/s
After calculations..
d = 5.68m

Now that we have distance, we can use this equation..
Δd= .5(v2 + v1)Δt
isolate for t and solve.
I got t = 1.075s

The rest should be a breeze.
Cheers!

PS: If you need help with the rest, just try first, and post your work and I'll see what I can do.