Question

the distance X along x-axis and height y for a body projected in x-y plane are given as X=6t and y=8t-3t². The initial speed of projection is (X,y in meters).​

Answer 1

Given:

The distance X along x-axis and height y for a body projected in x-y plane are given as x=6t and y=8t-3t².

To find:

Initial speed of Projection ?

Calculation:

$\rm \therefore \: x = 6t$

$\rm \implies \: v_{x} = \dfrac{dx}{dt}$

$\rm \implies \: v_{x} = \dfrac{d(6t)}{dt}$

$\rm \implies \: v_{x} = 6 \: m/s$

So, x component of initial velocity is 6 m/s.

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$\rm \therefore \: y = 8t - 3 {t}^{2}$

$\rm\implies \: v_{y} = \dfrac{dy}{dt}$

$\rm\implies \: v_{y} = \dfrac{d(8t - 3{t}^{2} )}{dt}$

$\rm\implies \: v_{y} = 8 - 6t$

$\rm\implies \: v_{y} \bigg|_{t = 0} = 8 - 6(0) = 8 \: m/s$

So, y component of initial velocity is 8 m/s.

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$\rm \therefore \: v_{net} = \sqrt{ {(6)}^{2} + {(8)}^{2} }$

$\rm \implies \: v_{net} = \sqrt{ 36 + 64 }$

$\rm \implies \: v_{net} = \sqrt{100 }$

$\rm \implies \: v_{net} = 10 \: m/s$

So, net initial velocity is 10 m/s.