Answer:
To achieve the velocity of 40 m/sec height will become 4 times
Explanation:
We have given initially truck is at rest and attains a speed of 20 m/sec
Let the mass of the truck is m
At the top of the hill potential energy is mgh and kinetic energy is 
So total energy at the top of the hill 
At the bottom of the hill kinetic energy is equal to
and potential energy will be 0
So total energy at the bottom of the hill is equal to 
Form energy conservation 
, for v = 20 m/sec

Squaring both side

h = 20.408 m
Now if velocity is 0 m/sec


h = 81.63 m
So we can see that to achieve the velocity of 40 m/sec height will become 4 times
Answer:
8.505 m
Explanation:
Let V1 and V2 be velocities of puck A and B respectively
Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s
Or
Vr=7.4 m/s
Distance=S= 18 m
Time =t=?
S=Vr×t
==> t=S/Vr
==> t= 18/7.4=2.43 sec
At this time both will strike together
<em><u>Distance by puck A</u></em>
<em>V1=3.5 m/s</em>
Time=t= 2.43 sec
Distance covered=d=?
d=V1×t=3.5×2.43=8.505 m
So, puck A will cover 8.505 meters before collision
If I’m understanding this. Th reason why the ball can resist gravity’s pull is because the force used to throw it up was greater than gravity’s pull on it. BTW, gravity’s pull is 9.8m/s. So it would have to be greater than that