Question

A body is projected down with velocity 5mspower-1 from height of 60 m its time of decents is

Answer 1

Given

A body is projected down with velocity 5 m/s from height of 60 m

To Find

Time ( of descent )

Concept Used

As the body is in free fall motion , so we need to apply equation of kinematics .

$\bf \pink{\bigstar\ \; s=ut+\dfrac{1}{2}at^2}$

where ,

• s denotes height
• u denotes initial velocity
• t denotes time
• a denotes acceleration

Time of descent denotes time taken to move from extreme to the ground

Solution

Initial velocity , u = 5 m/s

Height , s = 60 m

Acceleration , a = 10 m/s

Under gravity

Apply 2nd equation of motion ,

$\bf \green{\bigstar\ \; s=ut+\dfrac{1}{2}at^2}\\\\\to \rm 60=(5)t+\dfrac{1}{2}(10)t^2\\\\\to \rm 60=5t+5t^2\\\\\to \rm 12=t+t^2\\\\\to \rm t^2+t-12=0\\\\\to \rm t^2-3t+4t-12=0\\\\\to \rm t(t-3)+4(t-3)=0\\\\\to \rm (t-3)(t+4)=0\\\\\to \bf t=3\ \; \&\ \; t=-4$

But never be negative , so t = 3 s

So , Time of descent = 3 s

Answer 2

Given :-

• A body is projected down with velocity$\:\sf5ms^{-1}$from height of 60 m

To Find :-

• Its time of decents

Solution :-

$\boxed{\sf s = ut + \dfrac{1}{2} at^{2}} \: \: (\bf 2nd \: equation \: of \: motion)$

Where,

• s = Height
• u = Initial Velocity
• t = Time taken
• a = Acceleration

We have,

• Height(s) = 60 m
• Initial Velocity(u) = 5 m/s
• Acceleration(a) = 10 m/s

Substituting the values we get,

$:\implies\sf60 = (5)t + \dfrac{1}{\cancel2}(\cancel{10})t^{2}$

$:\implies\sf60 = 5t + 5t^{2}$

$:\implies\sf12 = t + t^{2}$

$:\implies\sf t^{2} + t - 12 = 0$

$:\implies\sf t^{2} - 3t + 4t - 12 = 0$

$:\implies\sf t(t-3) + 4(t-3) = 0$

$:\implies\sf (t - 3) (t + 4) = 0$

$:\implies\sf\underline{\boxed{\blue{\sf t = 3}}} \: and \: \underline{\boxed{\blue{ \sf t = - 4}}}$

We also know that,

• Time ( never be negative )

$\green{\therefore\sf The \: time \: of \: descent = 3 \: s}$

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Additional Information :-

★ 1st equation of motion :-

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

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time taken

★ 3rd equation of motion :-

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

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Covered

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