Question

A ball is dropped from roof of the tower of height h. The total distance covered by it in the last second of its motion is equal to the distance covered by it in first three second. The value of h meteer is

Answer 1

Answer:

125m

Explanation:

The distance covered in first 3 second

s=1/2gt2

=1/2×10×32

=45

If the ball takes 'n' second to fall to ground . The distance covered in nth second

sn=u+g/2(2n−1)

=0+10/2(2n−1)

=10n−5

Therefore 45=10n−5

n=5

Therefore h=12gt2

=12×10×25

=125m

Answer 2

Answer:

⇒ h = 122.5m

Explanation:

Given :

A ball is dropped from roof of the tower of height h.

The total distance covered by it in the last second of its motion is equal to the distance covered by it in first three second.

The value of h meter is ?

Solution :

Statement 1 :

A ball is dropped from roof of the tower of height h.

⇒ Total height = h

Statement 2 :

The total distance covered by it in the last second of its motion is equal to the distance covered by it in first three second.

⇒ Accerleration = acceleration due to gravity = 9.8 m/s²

⇒ total distance traveled in 1st 3 seconds of the motion = distance traveled in last 1 second.

∴ Let the total time taken to drop from height 'h' to ground be t,.

⇒ Total distance traveled in t seconds = h = ut + ¹/₂ at²

Distance traveled in 1st 3 seconds = x = 0(3) + ¹/₂ (9.8)(3)²

= 4.9 × 9 = 44.1 m

Velocity at t th second

⇒ v = u + at

⇒ u = a (t - 1)   (initial velocity at t th second , so Time (t) = (t - 1) )

⇒ Distance traveled in last t th second = x = a(t - 1)(1) + ¹/₂ a(1)²

⇒ 9.8 ( t - 1 ) + 4.9 (1) = 44.1

⇒ 4.9 [ 2(t - 1) + 1 ] = 4.9(9)

⇒ 2(t - 1) + 1 = 9

⇒ 2t - 1 = 9 ⇒ 2t = 10

⇒ $t = 5$

⇒ By substituting value of t ,

We get,

⇒ h = ut + ¹/₂ at²

⇒ h = (0) + ¹/₂ (9.8)(5)²

⇒ h = $4.9 (25)$

⇒ h = $122.5$

⇒ h = 122.5 m