A body. dropped from a height h at time t=0 reaches the ground at tune t0. would have reached a height h/2 at time:

Question

A body. dropped from a height h at time t=0 reaches the ground at tune t0. would have reached a height h/2 at time: (A)  (t0/2)  (B)  (t0/√2)  (C)  t02  (D)  (1/t02)

Tardigrade Admin · Accepted Answer

We have, h=(1/2)gt02 ...(i) Let t1 be the time taken by the body to reach a / height (h/2). Then (h/2)=(1/2)gt12 ⇒ (1/4)gt02=(1/2)gt12 [from (i)] ⇒ t1=(t0/√2)

Q. A body. dropped from a height $h$ at time $t = 0$ reaches the ground at tune $t_{0} .$ would have reached a height $h /2$ at time:

$\frac{t _{0}}{2}$

$\frac{t _{0}}{2}$

$t_{0}^{2}$

$\frac{1}{t _{0}^{2}}$

We have, $h = \frac{1}{2} g t_{0}^{2}$ ...(i)
Let $t_{1}$ be the time taken by the body to reach a / height $\frac{h}{2} .$ Then
$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$
$\Rightarrow$ $\frac{1}{4} g t_{0}^{2} = \frac{1}{2} g t_{1}^{2}$ [from (i)]
$\Rightarrow$ $t_{1} = \frac{t _{0}}{2}$

Q. A body. dropped from a height h at time t=0 reaches the ground at tune t0​. would have reached a height h/2 at time:

2t0​​

2​t0​​

t02​

t02​1​

We have, h=21​gt02​ ...(i) Let t1​ be the time taken by the body to reach a / height 2h​. Then 2h​=21​gt12​ ⇒ 41​gt02​=21​gt12​ [from (i)] ⇒ t1​=2​t0​​

Q. A body. dropped from a height $h$ at time $t = 0$ reaches the ground at tune $t_{0} .$ would have reached a height $h /2$ at time:

$\frac{t _{0}}{2}$

$\frac{t _{0}}{2}$

$t_{0}^{2}$

$\frac{1}{t _{0}^{2}}$

We have, $h = \frac{1}{2} g t_{0}^{2}$ ...(i)
Let $t_{1}$ be the time taken by the body to reach a / height $\frac{h}{2} .$ Then
$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$
$\Rightarrow$ $\frac{1}{4} g t_{0}^{2} = \frac{1}{2} g t_{1}^{2}$ [from (i)]
$\Rightarrow$ $t_{1} = \frac{t _{0}}{2}$