Question

The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y=(8t-5t^2)$ meter and x = 6t meter, where t is in second. The velocity with which the projectile is projected is

• A. $8m/sec$
• B. $6m/sec$
• C. $10m/sec$
• D. Not obtainable from the data.

Answer 1

The correct option is C $10m/sec$

Correct Option[c]

$\begin{array}{cc}x=6tm& ,y=(8t-5t{)}^{2}m\\ V_x=\frac{dx}{dt}& \Rightarrow V_x=\frac{d}{dt}\left(6t\right)\\ & =6{ms}^{-1}\\ V_y=\frac{dy}{dt}\Rightarrow & v_y-\frac{d}{dt}(8t-5t^2)\\ & =(8-10t){ms}^{-1}\end{array}$

At the time of projection, $t=0s$

$\begin{array}{c}\text{i.e.}\begin{array}{cc}{V}_y=& 8-10t\\ & =8-10\times 0=8{ms}^{-1}\end{array}\\ \begin{array}{c}V=\sqrt{V_{x^2}+V_{y^2}}\Rightarrow \sqrt{6^2+8^2}\\ V=\sqrt{36+64}\Rightarrow \sqrt{100}\end{array}\\ V=10{ms}^{-1}\end{array}$