Help on these Physics questions

Test Preparation AP Test Preparation
1

<p>ok, there’s a lot of questions, so even if u guys could help me with just 1, i’d appreciate it. btw, this was my quiz and i wasn’t sure on these answers so i want to know what the right ones are.</p>

<li><p>a rock is thrown straight up with an initial velocity of 19.6 m/s. what time interval elapses between the rock’s being thrown and its return to the original launch point?</p></li>
<li><p>a bird, accelerating from rest at a constant rate, experiences a displacement of 28 m in 11s. what is the avg. velocity?</p></li>
<li><p>mike throws a rock down with speed 14 m/s from the top of a 30 m tower. if g=9.8 m/s^2 and air resistance is negligible, what is the rock’s speed just as it hits the ground?</p></li>
</ol>

<p>4.maria throws 2 stones from the top edge of a building with a speed of 20 m/s. she throws one straight down and the other straight up. the first one hits the street in a time t1. how much later is it before the second stone hits? if there is not enough information, state it.</p>

<li>norma releases a bowling ball from rest; it rolls down a ramp with constant acceleration. after half a second it has traveled 0.75 m. how far has it traveled in 2 seconds?</li>
</ol>

2

<ol>
<li>avg v = (xf - xi) / t. avg velocity = (28-0) / 11 = 28/11</li>
</ol>

3

<p>got 1 and 2. just need 3, 4, and 5 now.</p>

4

<ol>
<li>vf^2 = vi^2 + 2a(delta y)
vf^2 = 14^2 + 2(9.8)(30)
vf^2 = 196 + 588 = 784
vf = 28.0 m/s</li>
</ol>

5

<ol>
<li>x=1/2at^2+v0*t+x0
.75=.5a(.5)^2
find a
a=6m/s^2
substite a in
x=1/2at^2 with t=2 seconds
x=1/2(6)(2)^2=12m</li>
</ol>

6

<p>
[quote]

4.maria throws 2 stones from the top edge of a building with a speed of 20 m/s. she throws one straight down and the other straight up. the first one hits the street in a time t1. how much later is it before the second stone hits? if there is not enough information, state it.

[/quote]
</p>

<p>In this example you have to realize conceptually that the final velocity of both stones right before they hit the ground is the same. That is, Vf2 = Vf1</p>

<p>Set up the two equations, one for stone 1 and one for stone 2. Knowing the above info, you realize that Vf's will not matter. You know Vi, a (obviously), and time is the variable, specifically for the second stone. Use the [Vf = Vi + at] equation.
Vf1 = Vi1 + a(t1)
Vf2 = Vi2 + a(t2)</p>

<p>Since you know that Vf1 = Vf2, set the two equations equal to each other.</p>

<p>Vi1 + a(t1) = Vi2 + a(t2)</p>

<p>Plug and chug...</p>

<p>-20 + (-9.8)(t1) = 20 + (-9.8)(t2)</p>

<p>They tell you the first stone hits the ground at time t1, so it's a known. Solve for t2.</p>

<p>-40 = (-9.8)(t2) - (-9.8)(t1)
-40/-9.8 = t2 - t1.
t2 = about 4 + t1.</p>