Question

d. An object thrown vertically upwards
reaches a height of 500 m. What was
its initial velocity? How long will the
object take to come back to the
earth? Assume g = 10 m/s
Ans: 100 m/s and 20 s​

Answer 1

Answer:

• initial velocity of object is 100 m/s
• time taken by object to come back to the earth (ground) is 20 seconds.

Explanation:

when object is thrown vertically upwards it reaches a height 500 m , then its final velocity will be zero , and acceleration due to gravity taken to be negative 10 m/s²

• final velocity, v = 0
• distance covered, s = 500 m
• acceleration due to gravity, g = - 10 m/s²

Let, initial velocity of object be u

then, Using third equation of motion

• 2 a s = v² - u²

[where s is distance covered, a is acceleration, v is final velocity , u is initial velocity of any body]

→ 2 g s = v² - u²

→ 2 (-10) (500) = (0)² - u²

→ -10000 = -u²

→ u = 100 m/s

therefore,

• initial velocity of object will be 100 m/s .

Now,

using first equation of motion

• v = u + a t

[ where v is final velocity, u is initial velocity, a is acceleration and t is time taken]

on covering 500 m upwards time taken by object, t₁ = ?

( initial velocity will be 100 m/s, final velocity will be zero , acceleration wil be negative )

→ v = u + a t₁

→ 0 = (100) + (-10) t₁

→ - 100 = -10 t₁

→ t₁ = 10 sec

Now,

using second equation of motion

• s = u t + 1/2 a t²

[ where s is distance covered, u is initial velocity, t is time taken , a is acceleration of body]

on covering 500 m downwards time taken by object, t₂ = ?

( while travelling downwards initial velocity of object will be zero, acceleration due to gravity will be positive )

→ s = u t₂ + 1/2 a t₂²

→ 500 = (0) t₂ + 1/2 (10) t₂²

→ t₂² = 100

→ t₂ = 10 sec

therefore,

total time taken by stone to reach ground will be,

→ t = t₁ + t₂

→ t = ( 10 ) + ( 10 )

→ t = 20 seconds

therefore,

• object will take 20 seconds to come back to the ground .