The equation of a projectile is y = ax - bx^2 | 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 $y = ax - bx^2$.When the projectile lands on the horizontal plane,its vertical displace

Vedclass · Accepted Answer

$\frac{a}{b}$

Vedclass · Answer

(A) The equation of the trajectory of a projectile is given by $y = ax - bx^2$.When the projectile lands on the horizontal plane,its vertical displacement $y$ becomes $0$.At this point,the horizontal distance $x$ is equal to the horizontal range $R$.Substituting $y = 0$ and $x = R$ into the equation,we get:$0 = aR - bR^2$$aR = bR^2$Since $R 
eq 0$ (as it is the range),we can divide both sides by $R$:$a = bR$$R = \frac{a}{b}$Thus,the horizontal range is $\frac{a}{b}$.

The equation of a projectile is $y = ax - bx^2$. Its horizontal range is ......

Similar Questions

The figure shows four paths for a kicked football. Ignoring the effects of air on the flight,rank the paths according to the initial horizontal velocity component,highest first.

In the given figure for a projectile,find the relation for $y$ in terms of $x_1, x_2$ and $\theta$.

$A$ water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $v$,the total area around the fountain that gets wet is :

$A$ stone is projected from the ground with velocity $50 \, m/s$ at an angle of $30^o$. It crosses a wall after $3 \, s$. How far beyond the wall will the stone strike the ground? $(g = 10 \, m/s^2)$

$A$ ball is projected from the ground with a speed $15 \, m/s$ at an angle $\theta$ with the horizontal such that its range and maximum height are equal. Then,$\tan \theta$ will be equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module