Phsyics Question?

Question

A projectile is shot upward from the surface of earth with an initial velocity of 120 meter per second. What is its velocity after 5 seconds? After 10 seconds?

The formula given is -4.9t^2 + v(subscript o)t + s(subscript o)

s(subscript o) is the initial height of the object and v(subscript o) is the initial velocity

Answer 1

Just substitute the appropriate numbers for variables.

In the equation, "t" is used to represent time. The initial velocity is given. The initial height is 0, since the object is on the surface at the beginning. By filling in our variable, we get:

For t = 5 seconds:
height = -(4.9 * 5^2) + (120 * 5) + 0 = 477.5 meters
For t = 10 seconds:
height = -(4.9 * 10^2) + (120 * 10) + 0 = 710 meters

Answer 2

4.9^2=24.01
24.01+120*5+height
24.01+610+height
634.01+height
height=634.01 meters...after 5 seconds
24.01+120*10+height
24.01+1200+height
1224.01+height
height after 10 seconds is 1224.01 meters
...just guessing

Answer 3

You need this equation:
a(or g)=(Vf-Vo)/t, where a=acceleration, or g=gravitation,
Vf= final velocity, Vo=initial velocity, t=time,a= -9.8 m/s^2
just sub. in, you will get the answer.
Vf= Vo+at=120m/s + (-9.8 m/s^2)x2s=100.4 m/s, after 2 seconds.
You can calculate it for 5 sec. and 10 sec.