physics
posted by Jon on .
A ball is thrown horizontally at 10.0 m/s from the top of a hill 50.0 m high. How far from the base of the hill would the ball hit the ground?
I don't understand the following explained by Quittich (sp)?
"Since there is no vertical component, the time to hit the ground is the same as if th ball had been dropped from a height of 50.0m. Calculate how long it takes for the ball to fall 50.0m. Take that time and multiply by the horizontal velocity (given as 10.0 m/s)".

The time that the ball is in the air just depends on how far it has to fall down to the ground. Since the ball was thrown HORIZONTALLY the initial VERTICAL velocity is 0. So, the ball immediately starts to drop. This happens at the same rate as if you just dropped it from the same height.
If you quickly rolled a ball off a table and then just as that ball rolled over the edge dropped another ball, they would both hit the ground at the same time because.
Back to the problem...
The time to hit the ground is then used to find out how far the ball traveled horizontally. The horizontal velocity is 10.0 m/s (given) and does not change. (Disregard air friction for this). So, the distance from the hill is just (time in the air) * (10.0 m/s).
Here is the formula to calculate the time in the air:
distance = (initial distance) + (initial velocity) * time + (0.5)*(acceleration)* (t^2)
Here, initial distance and initial velocity are both 0. So the equation becomes:
50.0m = (0.5)*(gravity)*(t^2)
Solve for t.
Then use the value you found for t.
horizontal distance = t * 10.0m/s 
t=1.02
initial velocity=10.0
I'm confused on what to do but from what you said I used the equation you gave me and plugged the numbers I thought were right and I can get an answer of 31.9 
that was before I knew you responded

OK, but when you get your answers, post them and we'll be sure they look good.

im just not getting anything. Im doing it 2 ways:
50.0m = (0.5)*(9.8)*(t^2)
4.9 = t
4.9*10=49
or 50.0m = (0.5)*(6.67 x 10^11)*(t^2)
3.335 x 10^11 = t
3.335 x 10^11 times 10.0 = I cant even get that 
OK, here are the steps to go through to solve the time eqution:
Just to be sure...
The expression t^2 means t to the second power or t * t.
Solving...
50.0m = (0.5)*(9.8 m/s^2)*(t^2)
50.0m = (4.9 m/s^2)*(t^2)
dividing both sides by (4.9 m/s^2)
(50/4.9)s^2 = t^2
10.2 s^2 = t^2
taking the square root of both sides
3.19s = t
So the horizontal distance is:
(3.19 s) * (10.0 m/s) = 31.9m 
ok. that's what I did it just didn't look right

Since there is no initial vertical velocity, gravity pulls the ball down 50 ft in t seconds which derives from h = Vot + gt^2/2.
SInce Vo = 0, 50 = 9.8t^2/2 making t = 3.19 sec.
In that 3.19 seconds, the ball travels horizontally a distance of d = 50(3.19) = 159.7m, ignoring air resistance. 
I think the horizontal velocity is 10.0m/s not 50.0m/s.