Question

16. An orange is thrown vertically upwards with
an initial velocity of 8 ms-1. (a) What will be the
velocity of orange when it returns to its starting
point? (b) How long will the orange take to reach
its starting point ?​

Answer 1

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Answer 2

Answer:

velocity of orange when it return to it's starting point = 8m/s in downward direction

t = 2*8/10 = 1.6sec

Explanation:

we can use newton's equation

v = u - at

s = ut - 1/2at^2

v^2= u^2 - 2as

here a = gm/s approx. 10m/s constant all over the world

here for time of flight s = 0 because s indicates displacement and displacement becomes zero when it reach to the ground.

ut = 1/2 gt^2

t = 2u/g

t = 2*8/10 = 1.6sec

now here v is the final velocity and u is the initially velocity

v^2 = u^2 - 2as

here further s is the displacement so it is zero when the orange reach to the ground

that is v^2 = u^2

so we can say that the orange will return to it's starting point with the same speed but different velocity

initially the velocity vector was in upward direction but at the time of hitting it's velocity vector is in downward direction

so velocity of orange when it return to it's starting point = 8m/s in downward direction

hope it is helpful.........................