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Motion in a Plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane5 Marks
Question
The height y and the distance x along the horizontal plane of a projectile on a certain plane (with no atmosphere) are given by $x = (6t)$ metre; $y = (8t - 5r^2)$ metre. where t is in seconds. What is the velocity with which the projectile is projected?
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Answer

$\text{x}=6\text{t}$$\therefore\ \frac{\text{dx}}{\text{dt}}=6\dots(\text{i})$
$\text{y}=8\text{t}-5\text{t}^2$
$\therefore\ \frac{\text{dy}}{\text{dt}}=8-10\text{t}\dots(\text{ii})$
Velocity with which it is projected can be written as:
$\text{v}^2=\Big(\frac{\text{dx}}{\text{dt}}\Big)^2+\Big(\frac{\text{dy}}{\text{dt}}\Big)^2$
Putting, t = 0 in equations (i) and (ii);
$\Big(\frac{\text{dy}}{\text{dt}}\Big)_{\text{t}=0}=8;\Big(\frac{\text{dx}}{\text{dt}}\Big)_{\text{t}=0}=6$
$\therefore\ \text{v}^2=6^2+8^2=36+64,\ \text{v}=10\text{m/s}$

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