Question

a body is projected vertically up with a velocity V and after some time it returns to the point from which it was projected. The average velocity and average speed of the body for the total time of flight are​

Answer 1

Answer:

10m per hour

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Answer 2

Explanation:

When body is projected against gravity the initial speed of projection is let say "u" and deceleration due to gravity is "g"

now by the formula of kinematics

v_f^2 - v_i^2 = 2 a sv f2

−v i2 =2as

now we will find the distance traveled by it till it stops

so we will have

0^2 - u^2 = 2(-g)s0 2−u 2=2(−g)s

s = \frac{u^2}{2g}s= 2gu 2

now we can say that total distance of motion will be double of this because here the distance it will move up will be same as the distance it will come back to ground again.

S = \frac{u^2}{g}S= gu 2

from above equation we can find the speed "u"

u = \sqrt{gS}u= gS

now the total time of flight can be found similar way

v_f = v_i + atv f=v i+at

when it reached to top point its velocity will become zero

0 = \sqrt{gS} - g t0= gS−gt = \frac{\sqrt{gS}}{g}t= ggS

t = \sqrt{\frac{S}{g}}t= gs

now the total time of flight will be double of this time as it will take same time to come back on ground

T = 2\sqrt{\frac{S}{g}}T=2 gS

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