Question

A projectile is fired from the surface of the earth with a velocity of 5 metre per second at an angle theta with the horizontal another projectile fired from the planet with a velocity of 3 metre per second at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the Earth. What is the value of the acceleration due to gravity on the planet?

PROJECTILE MOTION
CLASS 11 ​

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\Large\bold{\boxed{\boxed{3.5\:m/s^2 }}}$

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$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

If they follow same trajectory,

their range must be same:

Range :

$\bf\bold{\frac{u^2sin2\theta}{g}=\frac{v^2sin2\phi}{g'}}$ - - - (1)

since it is given(same angle) $\bf\bold{\theta=\phi}$

From equation 1

$\bf\bold{\frac{u^2sin2\theta}{g}=\frac{v^2sin2\theta}{g'}}$

Given v = 3 m/s , u = 5 m/s

$\bf\bold{\Rightarrow\frac{g'}{g}=\frac{v^2}{u^2}=\frac{9}{25}}$

$\bf\bold{\Rightarrow\:g'= g \times \frac{9}{25}=9.8 \times \frac{9}{25}}$

$\bf\bold{\Rightarrow \:g'=3.5\:m/s^2}$

Hence, the value of acceleration due to gravity is 3.5 m/s²

Answer 2

Answer:

${\green{\sf{\therefore acc^{n}\:of\:projectile\:on\:planet=3.6\:m/s^{2}}}}$

Explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

• In the given question information given about a projectile that is fired from earth surface and from a planet.

• Velocity of both the projectile is given and horizontal angle is same in both the condition.

• We have to find the acceleration of projectile on planet.

$\underline \bold{Given : } \\ \bold{From \: earth}\\ \implies u_{1} = 5 \: m /s \\ \implies {acc}^{n} = g_{1} = 10 \: m / {s}^{2} \\ \implies horizontal \: angle = \theta \\ \bold{for \: planet} \\ \implies u_{2} = 3 m /s \\ \implies horizontal \: angle = \theta \\ \\ \underline \bold{To \: Find : } \\ \implies {acc}^{n} = g_{2} = ?$

• According to given question :

• They have same trajectory. So, their range are equal.

$\bold{For \: earth : } \\ \implies Range = \frac{ { u_{1} }^{2} \sin2 \theta }{ g_{1}} \\ \implies Range = \frac{ ({5})^{2} \sin2 \theta }{10} \\ \implies Range = \frac{25 \sin2 \theta }{10} \\ \bold {\implies Range = 2.5 \sin2 \theta - - - - - (1)} \\ \\ \bold{For \: planet :} \\ \implies Range = \frac{ { u_{2} }^{2} \sin2 \theta }{ g_{2}} \\ \implies Range = \frac{ ({3})^{2}sin 2\theta }{ g_{2}} \\ \bold {\implies Range = \frac{9 \: sin2 \theta}{ g_{2} } - - - - - (2) } \\ \\ \bold {Comparing \: (1) \: and \: (2) } \\ \implies 2.5 \:sin2 \theta = \frac{9 \: sin2 \theta}{ g_{2} } \\ \implies g_{2} = \frac{9 \: sin2 \theta}{2.5 \: sin2 \theta} \\\bold{\implies g_{2} = 3.6 \: m /{s}^{2} } \\ \\ \bold {\therefore {acc}^{n} \: of \: projectile \: on \: planet = 3.6 \: m | {s}^{2} }$