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Grade 11Mechanics

The equation of the path of the projectile is y=0.5x-0.04x^2.The initial speed of projectile is
A)10m/s
B)15m/s
C)12.5m/s
D)None

jiten

8 Years agoGrade 11

2 Answers

Eshan

8 Years ago

Dear student,

The equation of path of a projectile is given as-

$y=xtan\theta-\dfrac{1}{2}\dfrac{gx^2}{u^2cos^2\theta}$
From the given equation, $y=0.5x-0.04x^2$

$tan\theta=0.5$
$\implies cos\theta=\dfrac{2}{\sqrt{5}}$

and $\dfrac{g}{2u^2cos^2\theta}=0.04$
$\implies u=\sqrt{\dfrac{g}{2(0.04)cos^2\theta}}=12.5m/s$

Khimraj

8 Years ago

from the equation of projectile

y = xtan $\Theta$ – ½(gx²)/(u²cos² $\Theta$ )

tan $\Theta$ = ½

so cos $\Theta$ = 2/ $\sqrt{5}$

and

g/2u²cos² $\Theta$ = 0.04

SO u = 12.5 m/s.

Hope it clears.

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