If the initial velocity in the horizontal dir | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The initial horizontal velocity is given as $u_x = 1\,m/s$.The equation of the trajectory is $y = 5x - 5x^2$.Differentiating the equation with res

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The initial horizontal velocity is given as $u_x = 1\,m/s$.The equation of the trajectory is $y = 5x - 5x^2$.Differentiating the equation with respect to time $t$,we get the vertical velocity component $v_y = \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$.First,find $\frac{dy}{dx} = \frac{d}{dx}(5x - 5x^2) = 5 - 10x$.Since $v_x = \frac{dx}{dt} = u_x = 1$ (assuming constant horizontal velocity),we have $v_y = (5 - 10x) \cdot 1$.At the initial point,$x = 0$.Therefore,the initial vertical velocity $u_y = (5 - 10(0)) = 5\,m/s$.Thus,the $y$-component vector of the initial velocity is $5\,\hat{j}$.

If the initial velocity in the horizontal direction of a projectile is the unit vector $\hat{i}$ and the equation of the trajectory is $y = 5x(1 - x)$,find the $y$-component vector of the initial velocity. (Take $g = 10\,m/s^2$) (in $,\hat{j}$)

Similar Questions

The horizontal range of a projectile is $4\sqrt{3}$ times its maximum height. Its angle of projection will be ......... $^\circ$.

$A$ particle is projected at an angle of $45^{\circ}$ with the horizontal with an initial kinetic energy $K$. What will be its kinetic energy at the highest point?

Neglecting air resistance,the time of flight of a projectile is determined by:

The maximum height attained by a projectile when thrown at an angle $\theta$ with the horizontal is found to be half the horizontal range. Then $\theta$ is equal to

Galileo,in his book $Two \text{ } new \text{ } sciences$,stated that "for elevations which exceed or fall short of $45^{\circ}$ by equal amounts,the ranges are equal". Prove this statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module