Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The height $y$ and the distance $x$ along the vertical 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:

8 m/sec

6 m/sec

10 m/sec

Not obtainable from the data

Step-by-Step Solution

Velocity Components: The velocity of a particle is defined as the time rate of change of its position. The components of velocity are given by $v_x = \frac{dx}{dt}$ and $v_y = \frac{dy}{dt}$ .
Differentiate Position Equations: Given $x = 6t$ , the horizontal velocity is $v_x = \frac{d}{dt}(6t) = 6 \text{ m/s}$ . Given $y = 8t - 5t^2$ , the vertical velocity is $v_y = \frac{d}{dt}(8t - 5t^2) = 8 - 10t \text{ m/s}$ .
Initial Velocity (at t=0): The projectile is projected at time $t = 0$ . Substituting $t=0$ into the velocity components: $v_{x0} = 6 \text{ m/s}$ . $v_{y0} = 8 - 10(0) = 8 \text{ m/s}$ .
Calculate Resultant Velocity: The magnitude of the initial velocity (speed of projection) is: $v = \sqrt{v_{x0}^2 + v_{y0}^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m/s}$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started