Question

a body dropped from certain height h reaches the ground after certain time t .after half of the time its height above ground is given 3h/4 bt hw ?explain?

Answer 1

a=9.8
t=t
u=0
s=ut+1/2at²
 =1/2at²=h
t=t/2 
H=1/2 X a X t²/4=h/4
height from ground=h-h/4=3h/4

Answer 2

Answer:

Height of the body after time t/2 above the ground is 3h/4.

Explanation:

Given that :

• Body dropped from height h reaches reaches ground after time t.

To find :

• Height of body from ground after time t/2.

Solution :

• Let, the body be dropped from point A which is h meter above the ground. After time t, it reaches the point B which is at the surface of ground.
• Distant travel by body during time t = h meters.
• Since, the body stopped from the rest it's initial velocity is zero.
• From second equation of motion

$s = ut + \frac{1}{2} gt² \\h = \frac{1}{2} gt² \\g = 2 \frac{h}{ {t}^{2} } ---(1)$

• Let, the distance travelled by the body during time t/2 be h' from point A. Then,

$h' = \frac{1}{2} g( { \frac{t}{2} }^{2} ) \\ = \frac{1}{2} \frac{2h}{ {t}^{2} } ( \frac{ {t}^{2} }{4} ) \\ h' = \frac{h}{4}$

• Distance travel by body during time t/2 from the point A is h/4
• Distance of the body from ground is
• h - h' = (h - h/4) = 3h/4.
• Hence, height of the body after time t/2 above the ground is 3h/4.