A)
I would say: the KE at the point of him grabbing it = the PE at his peak(all his KE has become PE) so KE=PE so ½mv²=mgh substituting:
.5*40*8²=40*9.8*h
1280=392h
h=3.265m
B)change 30 for 40 above and v is unknown
.5*30*v²=30*9.8*3.265
15v²=959.91
v²=63.994
v=7.999m/s
yes, it is...
Can someone get me started on this, I know how to work a diverging lens and a concave mirror, just not together. The problem:
A concave mirror with a radius of curvature of 20.0 cm is placed 25.0 cm from a diverging lens with a focal length of 16.7 cm. An object is placed midway between the...
And I know of the pain that you feel, the same as me
And I dream of the rain as it falls upon the leaves
And the cracks is our lives
Like the cracks upon the ground
They are sealed and now are washed away...
It was Iron Maiden - Rainmaker
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Guess this one:
I declare I don't...
doh, I'm an idiot. Using the same method as below, with .4 as the mass: So 7.33/.4 = 18.333 which is a, but minus gravity, the acceleration of the rocket is 8.53. That means it was at 38.4m and going 25.6 m/s when the engines cut off. Therefore it was 71.837 m high at the peak. Correct?
Did I solve this correctly?
A 0.4-kg model rocket is launched straight upwards. Its engine provides an impulse of 22 N*s by firing for 3 seconds. What is the average acceleration of the rocket while the engine is firing, and what is the maximum height the rocket will reach?
J=Ft so 22=F(3)...