Question

an object projected vertically upwards with an initial velocity 40 ms‐¹ from the ground maximum height reach the body​

Answer 1

u= 40ms^-1

v= 0ms^-1 (final velocity at maximum height is zero)

Now,

v^2= u^2 -2gh

0^2=(40)^2-2×10×h (g=10ms^-2)

20×h=1600

h=80m

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Answer 2

GIVEN :-

INITIAL VELOCITY = 40m/s

FINAL Velocity =0m/s (SINCE FINAL VELOCITY BECOMES ZERO WHEN AT MAXIMUM HEIGHT THE BODY LOSES ITS MOMENTUM)

ACCELERATION ACTING ON BODY = g =9.8m/s

NOW ACCORDING TO THIRD Equation OF MOTION.

SQUARE OF FINAL VELOCITY US EQUAL TO SQUARE OF INITIAL VELOCITY + DOUBLE OF PRODUCT OF ACCELERATION AND DISTANCE.

v^2 = u^2 +2as.

AFTER INPUTTING THE VALUES IN THIS EQUATION WE GET.

0^2 = 40^2 + 2(-9.8)s

0 - (-19.6)= 1600s

19.6 = 1600s

ON SHIFTING s TO L. H. S WE GET:

s = 1600/19.6

s= 81.6 m.

THEREFORE THE DISTANCE TRAVELED BY OBJECT IS 81.6m.

_______________________________

BY SCIVIBHANSHU

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