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 ​

Answer 1

Answer:

9/25 times of the gravity of earth

hope it helps

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{ acceleration\:of\:projectile\:on\:planet=3.6\:m/s}}^{2}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• 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.

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

• According to given question :

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

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