Question

A ball is dropped from a height of 45m .Then distance covered by it in last second of its motion will be

Answer 1

Given :-

Height of the ball dropped = 45 m

Initial velocity of the ball = 0 m/s

To Find :-

The distance covered by it in last second of its motion.

Analysis :-

Here we are given with the height, initial velocity and gravity.

Firstly find final velocity using the third equation of motion.

Then you can find the acceleration using the first equation of motion.

Finally, substitute the values we got accordingly using the second equation of motion and find the distance covered by it in last second of its motion accordingly.

Solution :-

We know that,

• v = Final velocity
• s = Displacement
• u = Initial velocity
• a = Acceleration

Using the formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2=u^2+2as}}$

Given that,

Initial velocity (u) = 0 m/s

Acceleration (g) = 10 m/s

Displacement (s) = 45 m

Substituting their values,

⇒ v² = 0² + 2 × 10 × 45

⇒ v² = 20 × 45

⇒ v² = 900

⇒ v = √900

⇒ v = 30 m/s

We know that,

• v = Final velocity
• t = Time
• a = Acceleration
• u = Initial velocity

Using the formula,

$\underline{\boxed{\sf First \ equation \ of \ motion=v=u+at}}$

Given that,

Final velocity (v) = 30 m/s

Initial velocity (u) = 0 m/s

Acceleration (g) = 10 m/s

Substituting their values,

⇒ 30 = 0 + 10 × t

⇒ 30 = 10t

⇒ t = 30/10

⇒ t = 3 sec

We know that,

• s = Displacement
• u = Initial velocity
• a = Acceleration

Using the formula,

$\underline{\boxed{\sf Second \ equation \ of \ motion=s=u + \dfrac{1}{2}a(2n-1)}}$

Given that,

Initial velocity (u) = 0 m/s

Acceleration (g) = 10 m/s

Time (t) = 10 sec

Substituting their values,

⇒ s = 0 + 1/2 × 10 × (2 × 3 - 1)

⇒ s = 1/2 × 10 × 5

⇒ s = 5 × 5

⇒ s = 25 m

Therefore, the distance covered by it in last second of its motion is 25 m.

Answer 2

Question :-

→ A ball is dropped from a height of 45m .Then distance covered by it in last second of its motion will be ?

Given :-

→ Height from where the ball is dropped = 45 m

→ Initial velocity of that ball = 0 m/s

To Find :-

→ Distance covered by it in last second of its motion.

Solution :-

→ As know that ,

$\sf\\The\;vertical\;distance\;covered\;by\;a\;freely\\\;falling\;body\;in\;time\;t\;is\;given\;by :-$

$\sf h\;=\;ut-\dfrac{1}{2}gt^{2}$

Where ,

h = vertical distance covered

u = initial velocity

t = time period

g = acceleration due to gravity = $\sf 10ms^{-2}$

→ Here , the object is dropped . So, the initial velocity is zero.

∴ The equation becomes :

 

$\sf h\;=\;-\dfrac{1}{2}gt^{2}$

→ Negative sign is used because we have assumed the origin height at 45 m from where the stone was dropped.

→ Now , let's solve the equation by substituting values !

$\sf - 45 = - \dfrac{1}{2} \times 10 \times t^{2} \\\\\\-45 = - \dfrac{10}{2} \times t^{2} \\\\-45 = -5 \times t^{2} \\\\t^{2} = \dfrac{-45}{-5} \\\\t^{2} = 9 \\\\t = \sqrt{9} \\\\t = 3 s$

→ Now , we need to find the distance covered

Δ$h_{t}_{-}_{1}_{,}_{t} \;=\; h_{t} - h_{t}_{-}_{1}$

 or

Δ$h_{t}_{-}_{1}_{,}_{t} \;=\; \dfrac{1}{2}g[t^{2} -(t-1)^{2} ]$

$\sf = \dfrac{1}{2} \times 10[3^{2} - 2^{2} ]\\\\\\= 25 m$

∴ The distance covered by it in last second of its motion is 25 m.

All done ! :D