Question

A ball is dropped from a height of 30 metre if its velocity increases uniformly at the rate of 10 metre per second square with what velocity will it strikes the ground calculate the time after which it strikes the ground

Answer 1

$\large\bf\underline{Given:-}$

• A ball is dropped from a height of 30 m and it's velocity increses uniformly at the rate of 10m/s²

$\large\bf\underline {To \: find:-}$

• Final velocity of ball
• the time after which it strikes the ground

$\huge\bf\underline{Solution:-}$

• Height (s) = 30m
• Initial velocity (u) = 0
• Acceleration (a) = 10m/s²
• Final velocity (v) = ?

Now finding the final velocity of ball by using 3rd equation of motion.

★Third equation of motion★:-

• v² = u² + 2as

↪ v² = 0² + 2 × 10 × 30

↪ v² = 20 × 30

↪ v² = 600

↪ v = √600

↪ v = 24.4m/s

★ Final velocity of ball = 24.4m/s

Now, Finding Time at which the ball strikes on ground using 2nd equation of motion.

★Second equation of motion★:-

• s = ut + ½at²

↪ 30 = 0 × t + ½ × 10 × t²

↪ 30 = 10t²/2

↪ 30 = 5t²

↪ t² = 30/5

↪ t² = 6

↪ t = √6

↪ t = 2.4s

Hence

≫ Final velocity of ball = 24.4m/s

Time at which the ball strikes on ground = 2.4s

$\rule{200}3$

✦ Some important concepts :-

Uniform velocity :- A body has a uniform velocity if it travels in a specific direction in a streight interval of time.

Acceleration : Acceleration of a body is defined as the rate of change of it's velocity with respect to time.

✦ Acceleration =$\tt\frac{change\:in\: velocity}{Time}$

★Equations of Motion :-

• v = u + at
• s = ut + ½× at²
• v² = u² + 2as

$\rule{200}3$

Answer 2

Answer:-

$\sf{The \ final \ velocity \ of \ ball \ will \ be \ 24.5 \ m}$

$\sf{and \ will \ strike \ the \ ground \ after \ 2.45 \ seconds.}$

Given:

• Height (s)=30 m

• Acceleration (a)=10 m/s^2

• Initial velocity (u)=0

To find:

• Final velocity (v)

• Time (t)

Solution:

$\sf{According \ to \ the \ third \ equation \ of \ motion.}$

$\boxed{\sf{v^{2}=u^{2}+2as}}$

$\sf{\therefore{v^{2}=0+2(10)(30)}}$

$\sf{\therefore{v^{2}=600}}$

$\sf{\therefore{v=\sqrt{600}}}$

$\sf{\therefore{v=10\sqrt6}}$

$\sf{\therefore{v=24.5 \ m \ s^{-1}}}$

$\sf{According \ to \ the \ first \ equation \ of \ motion.}$

$\boxed{\sf{v=u+at}}$

$\sf{\therefore{10\sqrt6=0+(10)\times \ t}}$

$\sf{\therefore{t=\frac{10\sqrt10}{10}}}$

$\sf{\therefore{t=\sqrt6}}$

$\sf{\therefore{t=2.45 \ seconds (approx)}}$

$\sf\purple{\tt{\therefore{The \ final \ velocity \ of \ ball \ will}}}$

$\purple{\tt{ be \ 24.5 \ m \ and \ will \ strike \ the}}$

$\sf\purple{\tt{ ground \ after \ 2.45 \ seconds.}}$