# Homework Help: Initial velocity of soccer ball and length of time in the air- projectile motion

1. Apr 15, 2012

### dani123

1. The problem statement, all variables and given/known data

A soccer ball is kicked at 38.0° with respect to the horizontal and travels 64.0m before striking the ground.
a) what is its initial velocity?
b) how long was it in the air?

2. Relevant equations

dv=1/2*at2
dh=Vh*Δt
Kinetic equation d=Vi*t+ 1/2*at2
dh=-V2*sin2θ /g
sinθ=opp/hyp
Δt=-2Vsinθ / g

3. The attempt at a solution

a) dh=cos(38)*64m= 50.4m
dv=sin(38)*64m=39.4m

Find t=√(39.4m/(0.5*9.8))=7.96s

plug this value into the d=Vi*t+ 1/2*at2 equation and solve for Vi if d=64m

Vi=3.14m/s

b) V=dh/Δt= 50.4m/7.96s=6.33m/s

Then solve for Δt=-2Vsinθ / g= 0.80s

I would like for someone to just double check my answers and that I used the appropriate equations and that the number of significant figures are being respected! Thanks so much in advance!

2. Apr 15, 2012

### BruceW

I don't think you've done this one correctly. When the question says it travels 64m before striking the ground, I think it means this is the horizontal displacement.

Edit: as a hint dh=Vh*Δt is the equation for the horizontal displacement, where you can use the horizontal component of initial velocity for Vh. And d=Vi*t+ 1/2*at2 is the equation for vertical displacement, where you can use the vertical component of initial velocity for Vi. You will need to make use of both these equations simultaneously to solve the problem.

Last edited: Apr 15, 2012
3. Apr 16, 2012

### dani123

Thank you for your time but Im still so lost here.

4. Apr 18, 2012

### BruceW

Maybe start with the equations that you need to use. The horizontal and vertical motion are separate, so there are 2 equations. Have you been through this kind of question in class?

5. Apr 22, 2012

### dani123

Well this a correspondance course that I taking through the department of education so Im kind of teaching myself, which is making these assignments a bit more challenging but thankfully I have this forum to kind of guide me in the right direction.

I tried to redo this problem after consulting my textbook again and came up with the following:
For part a) I used dh=-V2sin2/g, which lead me to getting V= 25.42m/s as my answer for this section.

For part b) Δt=Δx/Vx, and from this equation I got 2.52 seconds.

Are these answers correct? Thank you so much for your time and help.

6. Apr 22, 2012

### BruceW

You've got part a) correct. But I get a different answer for part b), what did you use for Δx and Vx ?

7. Apr 23, 2012

### dani123

I used Δx=64m and Vx=25.42m/s

8. Apr 23, 2012

### BruceW

Vx is the component of velocity in the horizontal direction, not the total initial velocity. So you need to do a little trigonometry to get Vx from V. (as a check, Vx should come out as less than V, so it will be less than 25.42m/s).

9. Apr 26, 2012

### dani123

Ok so I did a little trigonometry and got the velocity in the horizontal direction to be equal to 20.03m/s... So with that, I plugged it back into my delta t equation to get 3.195 or 3.20 seconds. Does this seem correct? Thanks!

10. Apr 26, 2012

### BruceW

Yep, that is all correct. Nice work! One other thing: I think you should only use 2 significant figures, in the answer, since you used 2 sig. fig. in the inputs of the equation (i.e. g=9.8 which is 2 sig. fig.)