Answer:
Explanation:
The trick is to find the time.
If the ball did not travel horizontally at all and was just dropped, how long would it take to hit the ground?
<em>Givens</em>
vi = 0
d = 1.2 m
a = 9.8 m/s^2
<em>Formula</em>
d = vi * t + 1/2 * a ^ t^2
<em>Solution</em>
1.2 = 0*t + 1/2 9.8 * t^2
1.2 = 4.9 t^2
t^2 = 1.2 / 4.9
t^2 = 0.245
sqrt(t^2) = sqrt(.245)
t = 0.4949 seconds
<em>Answer</em>
You are not very clear about which answer you want. There are two of them. One is going horizontally and the other is vertically.
<u>Vertical</u>
vf^2 = vi^2 + 2*a*d
vi = 0
a = 9.8
t = 0.4949
vf^2 =0^2 + 2 * 9.8 * 1.2
vf^2 = 23.52
sqrt(vf^2) = sqrt(23.52)
vf = 4.85 m/s
<u>Horizontal</u>
This is likely what you are looking for.
There is 0 acceleration horizontally.
d = 13.5 m
t = 0.4949
s = ?
s = d/t
s = 13.5/0.4949
s = 27.28
<em>Note</em>
You should note a couple of things.
- The horizontal and vertical speeds are not the same.
- The time is used for both the horizontal and vertical speeds.
- The horizontal and vertical distances are quite different.
- Horizontal accelerations for these type of questions is generally 0.
