Search any question & find its solution
Question:
Answered & Verified by Expert
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 $y=\left(8 t-5 t^2\right)$ meter and $x=6 t$ meter, where ${ }^t$ is in second. The velocity with which the projectile is projected is
Options:
Solution:
2651 Upvotes
Verified Answer
The correct answer is:
10 m/sec
$v_y=\frac{d y}{d t}=8-10 t \quad v_x=\frac{d x}{d t}=6$
at the time of projection i.e. $v_y=\frac{d y}{d t}=8$ and $v_x=6$
$\therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{6^2+8^2}=10 \mathrm{~m} / \mathrm{s}$
at the time of projection i.e. $v_y=\frac{d y}{d t}=8$ and $v_x=6$
$\therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{6^2+8^2}=10 \mathrm{~m} / \mathrm{s}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.