# Thread: Physics, vertical kinematics, gravity

1. ## Physics, vertical kinematics, gravity

I'm having trouble with a few problems... =(, they involve vertical movement

And these are equations that are supposed to help and use in each problem:

For the following eq.'s, Gravity = 9.8 m/s^2 or 32.2 ft/s^2

A) Final velocity = initial velocity + gravity x (time)
B) Displacement = 1/2[inital + final velocity]time
C) Displacement = [Initial velocity x time] + 1/2(gravity)(time^2)
D) [Final velocity]^2 = [Initial velocity]^2 + 2(gravity)(displacement)

1) 2 identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 30m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B is hits the ground does pellet A hit the ground?

When A reaches back to where it started and falls down, it has the same speed as B (30m/s) but like the problem states, it takes the trip up as well..

2) A wrecking hall is hanging at rest from a crane when suddenly the cable breaks. The time it takes for the ball to fall 1/2 way to the ground is 1.2 seconds. Find the time it takes for the ball to fall from the rest all the way to the ground

The answer is 1.7 seconds, and im totally lost on this one

2. And for the displacement, it is actually y, but y = displacement, or represents it

3. 1. $\displaystyle v_f = v_0 + at$

for gun A, let $\displaystyle t_1$ be the time for the pellet to hit he ground ...

$\displaystyle v_f = 30 - 9.8t_1$

for gun B, let $\displaystyle t_2$ be the time for the pellet to hit he ground ...

$\displaystyle v_f = -30 - 9.8t_2$

since $\displaystyle v_f$ is the same for both guns ...

$\displaystyle 30 - 9.8t_1 = -30 - 9.8t_2$

$\displaystyle 30 + 30 = 9.8t_1 - 9.8t_2$

$\displaystyle 60 = 9.8(t_1 - t_2)$

$\displaystyle 6.12 = t_1 - t_2$

2. $\displaystyle y = \frac{1}{2}at^2$

falling halfway ...

$\displaystyle \frac{y}{2} = \frac{1}{2}a(1.2)^2$

falling the entire way ...

$\displaystyle y = \frac{1}{2}at^2$

double the halfway equation ...

$\displaystyle y = a(1.2)^2$

since $\displaystyle y = y$ ...

$\displaystyle \frac{1}{2}at^2 = a(1.2)^2$

solve for t.

4. so i can substitute 9.8 m/s^2 for a ?

5. so i can substitute 9.8 m/s^2 for a ?
Yes, you can, but you won't need to. a will disappear when you solve for t

6. Originally Posted by realintegerz
Gravity = 9.8 m/s^2 or 32.2 ft/s^2