Question

Find the velocity with which the body will strike the ground if dropped from a height of 75.4m.Find also the time taken by it to reach the ground​

Answer 1

Given :

• Height of the tower = 75.4 m

To find :

• Time Taken to reach the ground.

• Velocity with which the body will strike the ground.

Solution :

To find the time taken to reach the ground :

We know the second Equation of Motion i.e,

$\bf{S = ut \pm \dfrac{1}{2}at^{2}}$

Where :

• S = Distance traveled
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

Now writing the second Equation of Motion when it is under gravity :

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

Where :-

• h = Height
• u = Initial Velocity
• t = Time Taken
• g = Acceleration due to gravity

From the first Equation of Motion :-

(When the initial velocity is 0) :

$:\implies \bf{h = ut \pm \dfrac{1}{2}gt^{2}}$

$:\implies \bf{h = 0 \times t \pm \dfrac{1}{2}gt^{2}}$

$:\implies \bf{h = \pm \dfrac{1}{2}gt^{2}}$

$\boxed{\therefore \bf{h = \pm \dfrac{1}{2}gt^{2}}}$

Hence the Equation formed for determining the height of a body is : ½gt².

Since , we are provided with the height, we can find the time taken . (Taking , g = 10 m/s²)

So using the above equation and substituting the values in it, we get :

$:\implies \bf{h = \pm \dfrac{1}{2}gt^{2}}$

[Since , the body is falling from the tower towards ground, i.e, under gravity, the acceleration due to gravity will be positive]

$:\implies \bf{h = \dfrac{1}{2}gt^{2}}$

$:\implies \bf{75.4 = \dfrac{1}{2} \times 10 \times t^{2}}$

$:\implies \bf{75.4 \times 2 = 10 \times t^{2}}$

$:\implies \bf{150.8 = 10 \times t^{2}}$

$:\implies \bf{\dfrac{150.8}{10} = t^{2}}$

$:\implies \bf{15.08 = t^{2}}$

$:\implies \bf{\sqrt{15.08} = t}$

$:\implies \bf{3.88(approx.) = t}$

$\boxed{\therefore \bf{t = 4\:s}}$

Hence, the time taken by the body to reach the tower is 4 s.

Velocity with which the body will reach the ground :

We know the First Equation of Motion. i.e, (Under gravity)

$\bf{v = u + gt}$

Where :

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration due to gravity
• t = Time Taken

Here , the initial velocity of the body will be 0 m/s.

Now using the first equation of the motion and substituting the values in it, we get :

$:\implies \bf{v = u + gt}$

$:\implies \bf{v = 0 + gt}$

$:\implies \bf{v = gt}$

$:\implies \bf{v = 10 \times 4}$

$:\implies \bf{v = 40}$

$\boxed{\therefore \bf{v = 40\:ms^{-1}}}$

Hence the velocity to reach the ground is 40 m/s.