Question

A ball gently dropped from a height of 20m . If it's velocity increases uniformly at a rate of 10m/s -1 with that velocity will it strike the ground? After what time will it strikes the ground.​

Answer 1

$v = \sqrt{2rg \: }$

$v = \sqrt{2 \times 20 \times 10 }$

V = 20m/s

$t = \sqrt{ \frac{2h}{g} }$

$t = \sqrt{ \frac{2 \times 20}{10} } = \sqrt{4} = 2$

t = 2 sec

Answer 2

Given :

• distance covered by ball, s = 20 m
• acceleration of ball, a = 10 ms⁻²

ball is gently dropped from a height therefore,

• initial velocity of ball, u = 0

To find :

• velocity at which ball strikes the ground, v = ?
• time after which ball strikes the ground, t = ?

Formulae required :

• third equation of motion

     2 as = v² - u²

• first equation of motion

     v = u + a t

( where a is acceleration , s is distance covered , v is final velocity , u si intial velocity , t is time taken )

Solution :

Using third equation of motion

→ 2 a s = v² - u²

→ 2 (10) (20) = v² - (0)²

→ 400 = v²

→ v = 20 ms⁻¹

therefore,

Ball will strike the ground with velocity 20 ms⁻¹.

Using first equation of motion

→ v = u + a t

→ 20 = 0 + (10) t

→ t = 20 / 10

→ t = 2 sec

therefore,

after 2 seconds ball will strike the ground .