Question

(d.) A stone thrown vertically upwards with
initial velocity u reaches a height 'h'
before coming down. Show that the
time taken to go up is same as the time
taken to come down.
leofaruddenly becomes twice
TE1
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Answer 1

We should not here that the distance between initial point (starting point) and final point of reach of the stone is constant for both upward and downward motion. So, we can say height for both upward and downward motion is same.

UPWARD MOTION CASE :

At highest point of reach of stone during upward journey, final velocity of stone (v) becomes zero.

Let the time taken for upward journey be "t".

Mathematical derivation :

v² - u² = 2 as

Or, 0² - u² = - 2gh

Or, h = u²/2g

Again, v = u + at

Or, 0 - u = - gt

Or, t = u/g

DOWNWARD MOTION CASE :

Height is same because the distance between final and initial point is same (constant) which is actually we are calling HEIGHT.

In this case, let the initial velocity be "U" and the final velocity be "V".

Height will be same (h).

Let the time taken for downward journey be "T".

And, for downward motion, acceleration due to gravity (g) will be positive and also the initial velocity (U) will be zero as it is falling downwards from resting point.

h = UT + 1/2 gT²

Or, u²/2g = 0 + 1/2 gT²

Or, u²/g = gT²

Or, u² = g²T²

Or, T = u/g

THUS, WE SAW THAT THE TIME FOR BOTH UPWARD AND DOWNWARD JOURNEY REMAINS SAME.

REMEMBER :-

(1) For upward motion acceleration due to gravity is negative because it is opposite to the direction of Earth's gravitational field.

(2) For downward motion acceleration due to gravity is positive because it is towards the direction of Earth's gravitational field.

(3) Height attained by the stone for upward journey = Height covered by stone for downward journey because the actual distance (height) is same between the two positions (initial and final) of stone.