First slide

Projectile motion

Question

The trajectory of a projectile in a vertical plane is $y = a x - b x^{2}$ , where a and b are constants and x and y are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are

Moderate

A

$\frac{b^{2}}{2 a}, \tan^{- 1} (b)$
B

$\frac{a^{2}}{b}, \tan^{- 1}$
C

$\frac{a^{2}}{4 b}, \tan^{- 1} (a)$
D

$\frac{2 a^{2}}{b}, \tan^{- 1} (a)$

Solution

$y = a x - b x^{2}; for height or y to be maximum:$
$\frac{d y}{d x} = 0 or a - 2 b x = 0 or x = \frac{a}{2 b}$
$i . y_{max} = a (\frac{a}{2 b}) - b {(\frac{a}{2 b})}^{2} = \frac{a^{2}}{4 b}$
$ii. {(\frac{d y}{d x})}_{x = 0} = a = \tan θ_{0}, where θ_{0} = angle of projection θ_{0} = \tan^{- 1} (a)$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App