Question

a ball is throw upward and it goes to the height 100m and comes down

1) what is the net displacement ?
2) what is the net distance ?
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Answer 1

Answer:

1. the net displacement is zeros because it came back to the initial position.
2. total distance covered by the ball = 100 + 100 = 200m

Answer 2

Explanation:

Acceleration ,a = -10m/s² (opposite direction of motion)

Final velocity ,v = 0m/s

Time taken ,t = 5s

To Find:-

Distance covered by the car before coming to rest ,s

Solution:-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀According to the Question

Firstly we calculate the initial velocity of the car . Using 1st equation of motion

v = u + at

where,

v is the final velocity

a is the acceleration

u is the initial velocity

t is the time taken

Substitute the value we get

:\implies:⟹ 0 = u + (-10) × 5

:\implies:⟹ -u = -50m/s

:\implies:⟹ u = 50m/s

So the initial velocity of the car is 50m/s

Now, calculating the final velocity of the car by using 3rd equation of motion.

v² = u² + 2as