Question

A particle is projected vertically up from the top of a tower with velocity 10m/s . It reaches the ground in 5 second find

Answer 1

Initial velocity is 10 m/s. Since the particle is projected up wards. The displacement would be 1/2 g*t^2-u*t. Therefore, the displacement would be 1/2 *10*25-50 which is equal to 75 m. I hope this is what u wanted.

Answer 2

Answer

• Height , h = 75 m
• Distance , d = 85 m

Correct Question

• A particle is projected vertically up from the top of a tower with velocity 10 m/s . It reaches ground in 5 s. Find height of tower and distance traversed by particle ?

Given

• Initial velocity , u = 10 m/s
• Total Time , t = 5 s

To Find

• Height of the tower , CD = ? m
• Distance traversed by particle , AD = ? m

Concept Used

Equations of motion

• v = u + at
• s = ut + ¹/₂ at²
• v² - u² = 2as

Solution

Case - 1 : AB

Initial velocity , u = 10 m/s

Final velocity , v = 0 m/s [ Since goes to rest ]

Acceleration , g = - 10 m/s² [ Against gravity ]

Distance , AB = s'

Apply 3rd equation of motion .

⇒ v² - u² = 2as

⇒ 0² - 10² = 2 ( -10 ) s'

⇒ - 100 = - 20 s'

⇒ s' = 5 m

Let time be t'

Apply 1st equation of motion .

⇒ v = u + at

⇒ 0 = 10 - gt'

⇒ 10t' = 10

⇒ t' = 1 s

Case - 2 : BD

Time , t" = t - t'

⇒ t" = 5 - 1

⇒ t" = 4 s

Acceleration , g = 10 m/s²

Initial velocity , u = 0 m/s [ ∵ After going to extreme position it starts fall down ]

Distance , BD = s"

Apply 2nd equation of motion .

s = ut + ¹/₂ at²

⇒ s" = ¹/₂ × 10 × (t")²

⇒ s" = 5(4)²

⇒ s" = 5(16)

⇒ s" = 80 m

So , Total distance traversed by particle , s = s' + s"

⇒ s = 5 + 80

⇒ s = 85 m

Height of the tower , h = CD

⇒ h = BD - BC

⇒ h = BD - AB [ ∵ AB = BC ]

⇒ h = s" - s'

⇒ h = 75 m