# Basic kinematics: projectile motion

Tags:
1. Nov 11, 2015

### Ryaners

This is a really basic problem but the answers I've gotten don't match up with the back of the book. I've been over it several times & can't understand where I'm going wrong - any pointers would be much appreciated!

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

A physics book slides off a horizontal tabletop with a speed of 1.40m/s. It strikes the floor in 0.320s. Ignore air resistance. Find a) the height of the tabletop off the floor; b) the horizontal distance from the edge of the table to the point where the book strikes the floor.

2. Relevant equations
vf = vi + at
x = xi + vit + ½at2

3. The attempt at a solution
I broke up the book's initial velocity into components:
Vf-x = Vi-x = 1.40m/s
Vf-y = Vi-y +at = -gt = -3.14m/s

The height of the table = the magnitude of the vertical displacement y (using the top of the table as the origin):
y = yi + Vi-yt + ½at2 = 0 + 0 + ½(-9.8m/s2)(0.320s)2= -0.502m -> height = 0.502m
What the book gives for this is 1.13m.

The horizontal distance = magnitude of horizontal displacement x:
x = xi + Vi-xt + ½at2 = 0 + (1.40m/s)(0.320s) + 0 = 0.448m
What the book gives here is 0.502m (which is what I got for the height - which means I'm either completely mixed up or maybe the book is wrong?!)

Thanks for taking a look.

Last edited: Nov 11, 2015
2. Nov 11, 2015

### haruspex

How did you get vfy=-4.9m/s? Looks like you used t=.5s, which, as it happens, would give a height closer to 1.1m.

3. Nov 11, 2015

### Ryaners

Yes, you're right, sorry! I've been trying other problems since getting stuck on this one & copied the wrong figure down from my rough work. It should be Vf-y = -3.136m/s

I'll edit that now.

4. Nov 11, 2015

### haruspex

Then I agree with your 0.5m height.