If the equation of motion of a projectile is | vedclass

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

Question

(A) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with the given equ

Vedclass · Accepted Answer

$\frac{A}{4}$

Vedclass · Answer

(A) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with the given equation $y = Ax - Bx^2$,we get:$A = 	an 	heta$ and $B = \frac{g}{2u^2 \cos^2 	heta}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.The range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.From $A = 	an 	heta$,we have $\sin 	heta = A \cos 	heta$.Substitute this into $B = \frac{g}{2u^2 \cos^2 	heta}$ to get $u^2 = \frac{g}{2B \cos^2 	heta}$.Now,$H = \frac{(\frac{g}{2B \cos^2 	heta}) (A^2 \cos^2 	heta)}{2g} = \frac{A^2}{4B}$.Also,$R = \frac{2 (\frac{g}{2B \cos^2 	heta}) (A \cos^2 	heta)}{g} = \frac{A}{B}$.The ratio of maximum height to range is $\frac{H}{R} = \frac{A^2 / 4B}{A / B} = \frac{A}{4}$.

If the equation of motion of a projectile is $y=Ax-Bx^2$,then the ratio of the maximum height reached and the range of the projectile is

Similar Questions

$A$ body is projected from the earth at an angle of $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \ m$, the maximum height reached by it is: (in meters)

The relation between the horizontal displacement $x$ (in metre) and the vertical displacement $y$ (in metre) of a projectile is $y = 3x - 0.8x^2$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ m/s^2$). (in $s$)

$A$ body is thrown horizontally from the top of a tower of height $5 \, m$. It touches the ground at a distance of $10 \, m$ from the foot of the tower. The initial velocity of the body is ......... $m/s$ $(g = 10 \, m/s^2)$.

What is the angle between the velocity and acceleration of a projectile at its maximum height (in $°$)?

Derive the formula for the range of a projectile $(R)$. Derive the formula for the maximum range of a projectile.

Vedclass Test Series

Exam Paper Generator

Online Exam Module