Science, asked by gaurbishakha, 9 months ago

a ball is dropped from a hieght of 20 m tower with 10 m/s2 rate of change velocity .what will be the velocity before its hit too the ground?

Answers

Answered by Anonymous

To Find :–

The Final velocity of the Ball before reaching the ground :-

Given :–

Height of the building = 20 m.

Acceleration of the ball = 10 m/s².

We Know :–

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ First' Equation of Motion :-

$\underline{\boxed{\bigstar\bf{v = u \pm gt}\bigstar}}$ .

Where :-

v = Final Velocity
u = Initial velocity
g = Acceleration due to gravity
t = time taken

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ Second Equation of Motion :-

$\underline{\boxed{\bigstar\bf{h = ut \pm \dfrac{1}{2}gt^{2}}\bigstar}}$

Where :-

h = Height of the Tower
u = Initial velocity
t = Time taken
g = Acceleration due to gravity

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ Third Equation of Motion :-

$\underline{\boxed{\bigstar\bf{v^{2} = u^{2} \pm 2gh}\bigstar}}$

Where :-

Note :- The Acceleration in laws of motion under Gravity is taken g.

Concept :–

According to the question, the ball is dropped from the top of the tower whose height is 20 m .

Hence, by the Using the formula for height , we can find the time taken to reach the ground and then by using the First Equation of Motion , we can find the final velocity of the ball.

Analysis :-

$\bullet$ The Acceleration due to gravity is taken as postive when the body is moving under Gravity.

$\bullet$ The Acceleration due to gravity is taken as negative when the body is moving against Gravity.

$\bullet$ The initial velocity is 0 when the body is falling .(u = 0)

$\bullet$ The Final Velocity is 0 at the highest point. (v = 0)

Obtaining the Equation from Second Equation of Motion , when u = 0.

Using the Second Equation of Motion and putting initial velocity as 0 in the Equation , we get :-

[Note :- The Acceleration due to gravity will be positive as the body is moving under Gravity ]

$:\implies \bf{h = ut + \dfrac{1}{2}gt^{2}} \\ \\ \\ :\implies \bf{h = 0 \times t + \dfrac{1}{2}gt^{2}} \\ \\ \\ :\implies \bf{h = \dfrac{1}{2}gt^{2}} \\ \\ \\ \therefore \purple{\bf{h = \dfrac{1}{2}gt^{2}}}$

Hence, the Formula obtained for height of an object is $\bf{h = \dfrac{1}{2}gt^{2}}$ .

Solution :–

To Find the Time Taken while Falling from the building :-

Given :-

Height = 20 m
Acceleration due to gravity = 10m/s²

Let the time taken be t sec.

Using the formula and substituting the values in it,we get :-

$:\implies \bf{h = \dfrac{1}{2}gt^{2}} \\ \\ \\ :\implies \bf{20 = \dfrac{1}{2} \times 10 \times t^{2}} \\ \\ \\ :\implies \bf{20 = 5 \times t^{2}} \\ \\ \\ :\implies \bf{\dfrac{20}{5} = t^{2}} \\ \\ \\ :\implies \bf{4 = t^{2}} \\ \\ \\ :\implies \bf{\sqrt{4} = t} \\ \\ \\ :\implies \bf{2 = t} \\ \\ \\ \therefore \purple{\bf{t = 2 s}}$ .

Hence, the time taken to reach the ground is 2 s.

To Find the Final velocity of the ball :-

Given :-

u = 0 m/s [Since it is falling under Gravity]

t = 2 s

g = 10 m/s²

Let the final velocity be v m/s

Using the First Equation of Motion and substituting the values in it ,we get :-

$:\implies \bf{v = u + gt} \\ \\ \\ :\implies \bf{v = 0 + 10 \times 2} \\ \\ \\ :\implies \bf{v = 10 \times 2} \\ \\ \\ :\implies \bf{v = 20} \\ \\ \\ \therefore \purple{\bf{v = 20 ms^{-1}}}$

Hence, the Final velocity is 20 m/s.

Alternative Method :-

Given :-

u = 0 m/s

g = 10 m/s²

S = 20 m

Let the Final velocity be v m/s .

Using the third Equation of Motion and substituting the values in it ,we get :-

$:\implies \bf{v^{2} = u^{2} + 2gh} \\ \\ \\ :\implies \bf{v^{2} = 0^{2} + 2 \times 10 \times 20} \\ \\ \\ :\implies \bf{v^{2} = 2 \times 10 \times 20} \\ \\ \\ :\implies \bf{v^{2} = 0^{2} + 2 \times 200} \\ \\ \\ :\implies \bf{v^{2} = 400} \\ \\ \\ :\implies \bf{v = \sqrt{400}} \\ \\ \\ :\implies \bf{v = 20} \\ \\ \\ \therefore \purple{\bf{v = 20 ms^{-1}}}$ .