Physics Kinematics Question (Check my work?)

[varyvod001]

Physics Kinematics Question! :) (Check my work?)

Homework Statement

A ball is thrown vertically up from a window that is 3.6 meters above the ground. The ball's initial velocity is 2.8 m/s up. What is the ball's velocity when it hits the ground? How long after the first ball should a second ball be dropped from the same window to ensure that both hit the ground at the same time?

Homework Equations

I used v_2^2=v_1^2 + 2aΔd & Δd = v_1Δt+ 1/2(aΔt^2) & d = 1/2 (v1+v2)t but the other kinematics equations may also be used.

The Attempt at a Solution

Rearranging the equation, (v_2^2 - v_1^2)/2a = Δd. So I plugged in v_2 as 0 to find where the ball would stop and start going down. A is gravity, -9.8 m/s. V_1 is 2.8. So I got -7.84/-19.6 which is 0.4 as the displacement. That means that the ball will have to travel that distance down to the window and then the 3.6 m down.
So now it's total down displacement is 4 m. Its v_1 is now 0. It's v_2 is what I want to find. I made down the positive direction because I didn't want to work with negatives so the displacement is 4 m and the acceleration is 9.8m/s^2. Plugging that into my equation, I got v_2 as being 8.85 m/s.

For the second part, I did the second mentioned equation for the second ball. I found that since v_1 was 0, that part could be removed from the equation and time was equal to 0.75 seconds.
The first ball would take 0.29 seconds to go up according to the 3rd equation and 0.91 seconds to come down according to the second.
0.29+0.91 = 1.2 and 1.2 - 0.85 = 0.35 so the second ball should be dropped 0.35 seconds later!
Any mistakes? :/

 

[dikmikkel]

Use energy conservation instead? It is relativley easy with that. Then divide the problem into 2 sections
1. Find the max height from the total energy with 0-potential at the groud:
$mgh_{window}+\dfrac{1}{2}mv_0^2 = mg(h_{max}+h_{window})$
$\dfrac{1}{2}mv_0^2 = mgh_{max}$
$h_{max} = \dfrac{v_0^2}{2g}$
So now the problem is to use energy conservation to answer the 1st q:
$mgh_{max} = 1/2mv_{hit}^2$
$v_{hit} = \sqrt{2gh_{max}}$
you can answer the 2nd question by demanding that the 2 objects has the same velocity and acceleration at the window(the same position is obvious) make your 2 equations and solve them for t( use that the velocity of the vertical thrown object is v0 when it goes down again, because of energy conservation)

 

[azizlwl]

1. correct
Taking downward as positive.
v²=u²+2as
v²=2.8²+2(9.8)(3.6)
v=8.85m/s

2. Correct
a. If the ball dropped,
t=√3.6(2)/(9.8)= 0.86sec.
b. Thrown up at initial speed of 2.8m/s
-3.6=2.8t-4.9t²
Using online calculator,http://www.math.com/students/calculators/source/quadratic.htm
and taking the positive root.
t=1.2sec

Time after 1st. ball thrown= 1.2-0.86= 0.34sec.