# Projectile motion of a flicked ball

1. Sep 30, 2011

### lulusmith

Really can't figure this out by dividing x and y. The only way i can do it is with equations that use quadratics and I can't do that in my head for a test were I'm not allowed a calculator.Please help, Thank you!!!

You flick a ball from a tabletop, so that the ball lands on the floor 1.20 m below. The ball's initial speed is 2.40 m/s. Use g = 9.80 m/s2.

(a) If the ball is launched off the table horizontally, what is the horizontal distance between the launch point and the point where the ball hits the floor?

(b) If the ball is launched off the table at a 30 degree angle above the horizontal, what is the horizontal distance between the launch point and the point where the ball hits the floor?

2. Oct 1, 2011

3. Oct 1, 2011

### lulusmith

So I looked at the site and this is what I did:
ok so I got part a.

I didn't get part b.
What I did was solve for t first:
Y=Yo+ VoyT + aT^2/2
Y= 1.2m + 1.2(got this from 2.4sin30)t + (-9.8)(t^2)/2
t+t^2 = -1.2x2/9.8
-1.24t = 1
t= 0.80

Then I used t to solve for ax:

Vx=Vox+at;
0= 2.08 (from 2.4cos30) + a(0.8);
a= 2.6

Then I used this to solve for x (AKA range):
x = Xo + VoxT +at^2/2
x= 2.08(0.8)+ 2.6(0.8^2)/2;
x= 2.50

But it was wrong, so what did I do wrong?

4. Oct 1, 2011

### mitch_1211

for the first part where you found t, you should use the quadratic formula to solve the expression y = ...

5. Oct 1, 2011

### lulusmith

Isn't there an easier way to solve it w/o using the quadratic formula?
(I'm not allowed to use a calculator during the exam)

6. Oct 1, 2011

### Staff: Mentor

You can solve it in steps:
- Find the time it takes to get to the highest point of its motion
- Find the distance of the highest point above the table (and the floor)
- Find the time it takes to fall to the floor from the highest point

Seems like quite a handicap not to use a calculator.

7. Oct 1, 2011

### lulusmith

Could you show me how to actually do the problem (e.g: the equation I would have to use)?

8. Oct 1, 2011

### Staff: Mentor

I'll give you the first step: v = v0 + at

What's the vertical speed at the highest point?

9. Oct 2, 2011

### mitch_1211

are you sure you're not allowed a calculator? That would make calculating the initial velocity in the x and y direction a bit difficult, unless they want you to leave it as 2.4sin30 or whatever