Question

A body thrown vertically upward from the ground reaches the maximum height in time 't'.The time taken by the body to reach the ground from that maximum height is equal to
(A)t/2
(B)t
(C)2t
(D)t^2​

Answer 1

Answer:

Option A

Explanation:

The time taken to reach h(max) will be the same as the time taken to reach ground from h(max) = t

But this can be proved logically...

Proof:

g is the acceleration due to gravity and h is the height.

When the ball is thrown up,

From 3rd equation of motion,

v² - u² = 2gh ...(i)

When thrown up:

The final velocity (v) of the body at h(max) = 0

Acceleration due to gravity = -g

= - u² = -2gh

u² = 2gh ...(ii)

On coming down:

Initial velocity (u) at h(max) = 0

Using (i),

v² = 2gh ...(iii)

From (ii) and (iii),

v² = u²

v = u (iv)

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Now, from 1st equation of motion,

v = u + at

Upward motion:

v = 0

a = -g

-u = -gt1

$\frac{u}{g} = t1$

Downward motion:

u = 0

a = g

v = gt2

$\frac{v}{g} =t2$

Since u = v (from iv) so $\frac{u}{g} = \frac{v}{g}$

Hence t1 = t2