# Homework Help: Projectile motion of ball and table

1. Dec 13, 2008

### sweedeljoseph

1. The problem statement, all variables and given/known data
If you launch a ball horizontally, moving at a speed of 2.24 m/s from a table that is 0.77 m tall, how far out from the base of the table will the launched ball land?

a) 0.001m b) 0.9m c) 3.17m d) 8.5m

2. Relevant equations
v=vo+at
v2=vo2+2ax
x=vot+1/2at2

3. The attempt at a solution
i COMPLETELY forgot how to do projectile thing. like i remember some of it but not fully. i know you do the table for x and y then solve for like v, vo, a, t, d all that good stuff. but i dont know if the speed that was given is for the x or y. im thinking the y. because if it was for x it would be the same for v and vo. someone please help i have finals on monday and tuesday!

thank you so much!!!
sweedeljoseph

2. Dec 13, 2008

### rl.bhat

When the ball leaves the table it has two velocities. One horizontal and other vertical. The horizontal velocity remains constant, because there is no acceleration in that direction. The vertical velocity changes, the acceleration due to gravity acts on it. Using your third equation, find the time it takes to reach the ground. Using this time, find the horizontal distance traveled by the ball.

3. Dec 13, 2008

### sweedeljoseph

i understand that. but is the speed given in the equation used as the horizontal velocity or the vertical velocity. thats where im getting stumped. i know if its horizontal it is the same for velocity and initial velocity.

4. Dec 13, 2008

### rl.bhat

In the problem, given velovity is horizontal velocity. Initially vertical velocity is zero. The vertical distance traveled by the ball is 0.77 m.