<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>1</u>
- Initial velocity=u=0m/s
- Final velocity=v=10m/s
- Time=10s=t




<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>2</u>
- initial velocity=0m/s=u
- Final velocity=v=0.25m/s
- Time=t=2s



Person-1 is accelerating faster.
Answer:
Law of conservation of momentum states that. For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.
Explanation:
Hope it helps
Answer:
Option C is the correct answer.
Explanation:
Considering vertical motion of ball:-
Initial velocity, u = 2 m/s
Acceleration , a = 9.81 m/s²
Displacement, s = 40 m
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
40 = 2 x t + 0.5 x 9.81 x t²
4.9t² + 2t - 40 = 0
t = 2.66 s or t = -3.06 s
So, time is 2.66 s.
Option C is the correct answer.
Answer:
Crumple zones are designed to absorb and redistribute the force of a collision. ... Also known as a crush zone, crumple zones are areas of a vehicle that are designed to deform and crumple in a collision. This absorbs some of the energy of the impact, preventing it from being transmitted to the occupants.
Answer:
a.18.5 m/s
b.1.98 s
Explanation:
We are given that

a.Let
be the initial velocity of the ball.
Distance,x=30 m
Height,h=1.8 m





Substitute the values





Initial velocity of the ball=18.5 m/s
b.Substitute the value then we get

t=1.98 s
Hence, the time for the ball to reach the target=1.98 s