# Homework Help: Projectile Motion with No Air Resistance

1. Nov 26, 2009

### 1irishman

1. The problem statement, all variables and given/known data
An object is thrown horizontally from the top of a cliff at a velocity of 20m/s.
If the object takes 4.20s to reach the ground, what is the range of the object?
I got the first question figured out, but i can't figure out the second question which is:
What is the velocity of the object when it hits the ground? This is the one I can't seem to figure out.

2. Relevant equations
v=d/t horizontal uniform motion equation
d= vit +1/2at^2 vertical uniformly accelerated motion equation

3. The attempt at a solution
I got the range to be 84m in terms of the time and velocity given for the answer to the first question that applies to the horizontal distance.

The final velocity when the object hits the ground will be zero I figure. The initial velocity on the vertical will be zero too right? Help?

2. Nov 26, 2009

### JaWiB

It probably means right before the object hits the ground

So you need to figure out the final x- and y- components of its velocity and then find the magnitude

3. Nov 26, 2009

### 1irishman

I'm not sure how to do that or what equation/s to use to be honest.
Is it true that time will be the same for both horizontal and vertical component? Will initial vertical velocity be equal to zero?
If I have t=4.20, initial vertical velocity=0, and a=9.80m/s^2
can I somehow use d=vit +1/2at^2 to get final velocity just before object hits ground? Or do I have to use Vf^2=vi^2+2ad if I can find height vertical which I think is 86m. Help?

4. Nov 26, 2009

### JaWiB

The problem states that it is thrown horizontally, so initially it has no velocity in the vertical direction.

Separate everything into horizontal and vertical components:
What is the net force in the horizontal direction? The acceleration?
Same for the vertical?

>Or do I have to use Vf^2=vi^2+2ad if I can find height vertical which I think is 86m.

Yes, if you have constant acceleration you can use that equation

5. Nov 26, 2009

### 1irishman

This is what I got:

vertical distance:
d=1/2at^2
1/2X9.8X4.2^2=86.2m
-------------------------
then final vertical velocity is:
0=20^2+2X9.8X86
=400+1686
=sqrt 2086=45.7m/s is final velocity before hits the ground
they have in the book 45.8m/s before hits the ground at 64.1deg below the horizontal
I had 45.7m/s at 66deg below horizontal tan-1 45.7/20

6. Nov 27, 2009

### 1irishman

Since Vx = 20 m/s, Vy = g*t = -9.80 * 4.20 = -41.2 m/s

Therefore, the speed of the object is |V| = sqrt ( Vx^2 + Vy^2 ) = sqrt ( (41.2)^2 + 20^2) = 45.8 m/s

The direction of velocity is : tan (theta) = Vy / Vx = -41.2 / 20 = - 2.06 ------> theta = tan-1( -2.06 ) = - 64.1 negative means below horizontal.
that's just so much easier!