If a projectile is thrown obliquely both at earth and moon and the angle of projection with the horizontal and the initial speed of throw is same in both earth and moon then which projectile will cover the maximum height?

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If a projectile is thrown obliquely both at earth and moon and the angle of projection with the horizontal and the initial speed of throw is same in both earth and moon then which projectile will cover the maximum height?

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Answer 1 · 2017-08-11T05:48:10

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Maximum height is reached when $sin θ$ $sin theta$ component of the velocity of projectile becomes 0 (zero)
Use the kinematic formula

$v^{2} - u^{2} = 2 g h$ $v^2-u^2=2gh$

Since force of gravity is acting against the direction of vertical component of initial velocity so it is deceleration.
Substituting various values

$0^{2} - {(u sin θ)}^{2} = 2 (- g) h$ $0^2-(usintheta)^2=2(-g)h$
$\Rightarrow h = \frac{u^{2} {sin}^{2} θ}{2 g}$ $=>h=(u^2sin^2theta)/(2g)$

Acceleration due to gravity on the Earth is $= 9.81 m s^{- 2}$ $=9.81ms^-2$ and acceleration due to gravity on the surface of the Moon is $= 1.625 m s^{- 2}$ $=1.625 ms^-2$

From the derived expression for maximum height we see that for same initial velocity and same angle of projection

$h \propto \frac{1}{g}$ $hprop1/g$

$\Rightarrow$ $=>$ Maximum Height attained on moon will be greater.

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If a projectile is thrown obliquely both at earth and moon and the angle of projection with the horizontal and the initial speed of throw is same in both earth and moon then which projectile will cover the maximum height?

1 Answer

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