Question

A football is kicked from the ground with a speed of 23m/s at an angle of 42 degrees above the horizontal

a) how long is the football in the air

b) how far does it travel

Answer 1

A) How long, i suppose that's the time
Time, T = 2uSin420†2/g
T = 2 x 23 x 0.6691/10
T = 31.37/10
T = 3secs.

B) I suppose that's the height too.
Maximum height, H = u0…5Sin0…5420†2/2g
H = 230…5 x (0.6691)0…5/2 x 10
H = 529 x 0.4477/20
H = 236.8333/20
H = 11.84m

Answer 2

First, break down the velocity into its' horizontal and vertical components:

Vx=23m/s*Cos42=17.09m/s
Vy=23m/s*Sin42=15.39m/s

Use the following kinematic equation:

Vf=Vi+at

Where

Vi=15.39m/s
a=g=-9.8m/s
t=???
and
Vf=0m/s
***When the football reaches it's peak, the velocity will be 0m/s

Solve for t using Vy for Vi:

0=15.39m/s+(-9.8m/s^2)t

-15.39m/s/-9.8m/s^2=t

t=1.57s

It takes the same amount of time for the ball to reach the top, as it takes to come back down. So, the total time is

2*t=2*(1.57s)=3.14s=3.1s

Joel and I agree.

I don't think that the author of the question want's to know the distance that the ball travels in the y-direction, but the author wants to know the distance in the horizontal direction. To solve for that, use the following equation:

d=Vx*t

d=17.09m/s*3.14s

d=53.67m=54m

Answer 3

Wow..thats correct Devron, i dint see that from that angle. Well that's why u're there.

Answer 4

A shot-putter launches a 7.5kg shot with a speed of 12m/s at an angle of 37 degrees above the horizontal. if, upon release, the shot was 2.0m above the ground, determine (a) the horizontal range of the shot. (b) the speed of the projectile through the air just before it hits the ground.