Question

A body dropped from the top of tower takes 4 sec to hit the ground.How much time does it takes to cover second half tower (g =10 m/s^2)​

Answer 1

Answer :

Time taken by body to hot the ground = 4s

We have to find time taken by body to cover second half of the tower$.$

★ First of all we need to find height of the tower.

For a body falling freely under the action of gravity, g is taken positive.

For free falling motion, u = zero

➝ H = ut + 1/2 gt²

➝ H = 0 + 1/2 (10)(4)²

➝ H = 1/2 (160)

➝ H = 80 m

Now let's find time taken by body to cover first half of the tower.

Height of first half will be 40m.

➝ H/2 = ut₁ + 1/2 gt₁²

➝ 40 = 0 + 1/2 (10)(t₁)²

➝ 40 = 5t₁²

➝ t₁² = 8

➝ t₁ = 2.82 m

Time taken by body to cover second half :

➝ t = t₁ + t₂

➝ 4 = 2.82 + t₂

➝ t₂ = 1.18 s

∴ Body will take 1.18s to cover second half of the tower.

Answer 2

Answer:

$\: \fbox{⛄given}$

Time = 4 second

g= 10 m/sec²

inital Velocity= 0 m/sec²

$\fbox {To find}$

time does it takes to cover second half tower

$\fbox{solution}$

♠️ Let the height of the tower be x.

Using the second equation of motion ,we will find the height of the tower.

$s = ut + \frac{1}{2} a {t}^{2}$

since the ball is dropped the inital Velocity will be 0.

x= 0 + 1/2 x 10 x 4²

x= 80 metre.

⭐Time taken to cover the first half of the tower....

height of the half tower = 80/2 = 40 metre

x/2= 0+ 1/2 x 10 x t²

40 = 0+ 1/2 x 10 x t²

t= 2.82 seconds

⭐Time taken to cover the second half of the tower-

T (total) = T1+ T2

T2= 4- 2.82

T2= 1.18 seconds

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