# TEST question

1. Nov 9, 2007

### pooface

TEST question!!

1. The problem statement, all variables and given/known data
Golf ball is struck at 12 m above the plain of a tee giving it a velocity of 40m/s at an angle of 30degrees.

a) find the time of flight
b) find max height above plane
c) horizontal distance to point of landing.
d) final velocity and angle.

2. Relevant equations

3. The attempt at a solution
Vox = 40cos30
Voy = 40sin30

a) -12 = 40sin30t + 0.5(-g)t^2
t=4.6084 seconds

b) disregarding the 12m elevation (which i will add later on).
0=40sin30t + 0.5(-g)t^2
t=4.0775s
t/2 = time at peak
y= - 0.5(-g)t^2 +12
= 32.387m

c)x =voxt
= 34.34(4.6084)
=158.25m

d)
-12= Vfy(4.6084) -0.5(-g)(4.6084)^2
Vfy = 25.21m/s
angle = arctan(vfy/vox) = -36.28deg

Last edited: Nov 9, 2007
2. Nov 9, 2007

3. Nov 9, 2007

### pooface

the a),b),c),d)

4. Nov 9, 2007

### catkin

That's not a question.

What do you want?

5. Nov 10, 2007

### pooface

What i meant was, confirmation of process.

Have i done this correctly? I want to check because I want to see if I screwed up a test question.
Thanks.

6. Nov 10, 2007

### Kurdt

Staff Emeritus
I'd have a look at b again. You can't assume that the maximum height comes at half the flight time because of the difference in elevation from where it was hit and where it lands.

I'd also try part d again. The velocity in the formula you're using there is an initial velocity.

7. Nov 10, 2007

### pooface

part b) what i did was I neglected the initial height because i want to calculate the time of impact then half that time. Using this time, I get the height. but i would have to add the 12m back to get the real height.

I also could have used

Vfy=Voy+at

0=-20+(-9.81)t

which equals the same 2.039s.

part d) since vox is constant the final would be the same as well. so i just thought i would still name it vox.

8. Nov 10, 2007

### pinkyjoshi65

Hey, I got a question. Isint this just direct applcations of projectile formulas.
Like: DeltaT= 2ViSintheta/g and so on?

9. Nov 11, 2007

### learningphysics

Everything looks right to me except b). d) is correct... but a couple of little things are there...

For b), they want the height above the plane... not the height above the ground... so you wouldn't add 12...

also, you want: d = v1*t + (1/2)at^2 = 40sin30t + 0.5(-g)t^2... plug in t = 4.0755/2 will work.. gives 20.408m

you can also use: v2^2 = v1^2 + 2ad. v2 = 0, v1 = 40sin(30). so d = 40^2[sin(30)]^2/(2g) = 20.408m

Kurdt, the formula being used in part d, is d = vf*t - (1/2)at^2 which is a correct formula.

In part d, your Vfy should be -25.21 which is probably what you meant. So the final speed is sqrt[(-25.21)^2 + (34.34)^2] = 42.6 m/s. and the angle is right -36.28 degrees. ie: 36.28 degrees below the horizontal

Last edited: Nov 11, 2007