# Physics, vertical kinematics, gravity

• Sep 17th 2008, 05:34 PM
realintegerz
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
• Sep 17th 2008, 07:40 PM
realintegerz
And for the displacement, it is actually y, but y = displacement, or represents it
• Sep 17th 2008, 07:58 PM
skeeter
1. $v_f = v_0 + at$

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

$v_f = 30 - 9.8t_1$

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

$v_f = -30 - 9.8t_2$

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

$30 - 9.8t_1 = -30 - 9.8t_2$

$30 + 30 = 9.8t_1 - 9.8t_2$

$60 = 9.8(t_1 - t_2)$

$6.12 = t_1 - t_2$

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

falling halfway ...

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

falling the entire way ...

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

double the halfway equation ...

$y = a(1.2)^2$

since $y = y$ ...

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

solve for t.
• Sep 17th 2008, 08:13 PM
realintegerz
so i can substitute 9.8 m/s^2 for a ?
• Sep 17th 2008, 08:48 PM
Quote:

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
• Sep 18th 2008, 09:25 AM
topsquark
Quote:

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