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.
<span>Kinetic
energy is the energy that is possessed by an object that is moving. It is
calculated by one-half the product of the mass and the square of the velocity
of the object. We need to determine the velocity of the cart that is moving.
</span>
First, we use Newton's Second Law of motion;
<span>Force = ma
200 = 55a
a = 3.64 m/s^2
Then, from the kinematic equation we calculate the velocity;
v^2 = v0^2 + 2ax
where v is the final velocity, v0 is the initial veocity (zero since it initially start at zero), a is the acceleration ( 3.64 m/s^2) and x is the distance traveled.
v^2 = 0^2 + 2 (3.64) (10)
v^2 = 72.73 m^2 / s^2
v = 8.53 m / s
KE = mv^2 / 2
KE = 1/2 (55) (8.53^2)
KE= 2000 J</span>
F=ma = 1000kg * 9.8 m/s^2 = 9800 N
