M = 600 kg
u = 0 (as the car was at rest initially)
v = 5 m/s
Initial kinetic energy = ½mu²
= ½×600×0
= 0
Final kinetic energy = ½mv²
= ½×600×25
= 7500 J
Answer:
4.14 m
Explanation:
In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s .
Let in this last leg , u be the initial velocity.
s = ut + 1/2 g t²
2 = .2 u + .5 x 9.8 x .04
u = 9.02 m /s .
Let v be the final velocity in this leg
v² = u² + 2 g s
v² = (9.02)² + 2 x 9.8 x 2
= 81.36 +39.2
v = 10.97 m / s
Now consider the whole height from where the ball dropped . Let it be h.
Initial velocity u = 0
v² = u² +2gh
(10.97 )² = 2 x 9.8 h
h = 6.14 m
Height from window
= 6.14 - 2m
= 4.14 m
Answer:
<h2>Solving elastic collisions problem the hard way</h2><h3 />
Explanation:
perfect drawing
