Equation of projectile is y = 16x, - ,frac{{5{x^2}}}{4} , horizontal range is ..........

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3-2.Motion in Plane

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The equation of projectile is $y = 16x\, - \,\frac{{5{x^2}}}{4}$, The horizontal range is .......... $m$

A

$16$

B

$8$

C

$3.2$

D

$12.8$

Solution

$y = x\tan \theta \left[ {1 – \frac{x}{R}} \right]$

$y = 16x – \frac{{5{x^2}}}{4}$

$y = 16x\left[ {1 – \frac{x}{{64/5}}} \right]$

$R = \frac{{64}}{5} =12.8 m.$

Standard 11

Physics

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