Question

a bullet fired vertically up from the ground reaches a height 40 m in in it's path from the ground and it takes further time 2 seconds to reach the same point during descent . The total time of flight is

Answer 1

T = 2v/g
2 = 2v/g
v = g
v = 9.8 m/s

v² - u² = 2aS
(9.8)² - u² = -2*9.8*40
u² = 9.8(9.8 + 80) = 880.04
u = 29.6 m/s

T = 2u/g
T = 2*29.6/9.8
T = 6.04 seconds

Answer 2

Given that,

Height = 40 m

Time = 2s

We need to calculate the final velocity

Using formula of time flight

$T=\dfrac{2v}{g}$

Where, T = time

g = acceleration due to gravity

Put the value into the formula

$2=\dfrac{2v}{9.8}$

$v=9.8\ m/s$

We need to calculate the initial velocity

Using equation of motion

$v^2-u^2=2gh$

Where, h = height

v = final velocity

Put the value into the formula

$(9.8)^2-u^2=-2\times9.8\times40$

$-u^2=-784-96.04$

$u=\sqrt{880.04}$

$u=29.6\ m/s$

We need to calculate the total time of flight

Using formula of time of flight

$T=\dfrac{2u}{g}$

Put the value into the formula

$T=\dfrac{2\times29.6}{9.8}$

$T=6.0\ sec$

Hence, The total time of flight is 6.0 sec.