A projectile is fired from the surface of the earth with a velocity of 5m s ^ - 1 and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3 ms ^ - 1 at the same angle follows a

Correct answer: A. 3.5 The Equation of the Trajectory of a Projectile y = xtan(θ) - g x ^ 2 2 u ^ 2 (θ) y ,x are Variable and θ is constant so g u ^ 2 = constant g 5 ^ 2 = a 3 ^ 2 a = 9.8 × (9)/(25) a = 3.5

NEET 2014

Exam: NEET
Year: 2014
Subject: Physics
Chapter: Motion in Two Dimensions

NEET 2014 · Physics · Motion in Two Dimensions

A projectile is fired from the surface of the earth with a velocity...

A projectile is fired from the surface of the earth with a velocity of

5m{s}^{- 1}

and angle

\theta

with the horizontal. Another projectile fired from another planet with a velocity of

3{ms}^{- 1}

at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is: (given

= 9.8m{s}^{- 2}

)

A projectile is fired from the surface of the earth with a velocity of 5m s ^ - 1 and angle θ with the horizontal.

Options

A $3.5$
B $5.9$
C $16.3$
D $110.8$

Correct answer

A. $3.5$

Step-by-step solution

A projectile is fired from the surface of the earth with a velocity of 5m s ^ - 1 and angle θ with the horizontal.

The Equation of the Trajectory of a Projectile

$y = xtan\left(\theta\right) - \dfrac{g{x}^{2}}{2{u}^{2}\cos \left(\theta\right)}$

y ,x are Variable and $\theta$ is constant

so $\dfrac{g}{{u}^{2}} = constant$

$\dfrac{g}{{5}^{2}} = \dfrac{a}{{3}^{2}}$
$a = 9.8 \times \dfrac{9}{25}$
$a = 3.5$

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