Question

a ball is thown vertically upwards at 50m/s.find the greatest height attained and the velocity after 2.5s​

Answer 1

Answer:

answer is as given in image

Answer 2

Answer

• Greatest height attained is 125 m
• velocity after 2.5 s is 25 m/s

Given

• A ball is thrown vertically upwards at 50 m/s

To Find

• The greatest height attained and the velocity after 2.5 s ​

Concept

Equation's of motion

• v² - u² = 2as
• v = u + at

where

• s denotes distance distance travelled
• u denotes initial velocity
• a denotes acceleration
• t denotes time

Solution

Initial velocity , u = 50 m/s

Final velocity , v = 0 m/s

[ ∵ Finally goes to rest at extreme position ]

Time , t = 2.5 s

Acceleration due to gravity , a = g = - 10 m/s²

[ ∵ Thrown against the gravity ]

Distance travelled , s = ? m

Velocity after 2.5 s , vₓ = ? m/s

________________________

Apply 1 st equation of motion .

⇒ v = u + at

⇒ vₓ = 50 + (-10)(2.5)

⇒ vₓ = 50 - 25

⇒ vₓ = 25 m/s

________________________

So , velocity after 2.5 s is 25 m/s

________________________

Apply 3 rd equation of motion .

⇒ v² - u² = 2as

⇒ 0² - 50² = 2(-10)(s)

⇒ - 2500 = - 20 s

⇒ 125 = s

⇒ s = 125 m

_____________________________

So , Greatest height attained is 125 m

_____________________________