Question

In: Physics

A projectile is launched vertically from the surface of the Moon with an initial speed of...

A projectile is launched vertically from the surface of the Moon with an initial speed of 1290 m/s. At what altitude is the projectile's speed one-fifth its initial value?

Expert Solution

Hey there!

First let me give you the formula for the Vertical distance and the vertical velocity of a projectile.

Where,
the velocity along the y-axis is V_y ,
the initial velocity along the y-axis is V_y1 .
acceleration due to gravity is g, and
the time taken is t.

The acceleration due to gravity in moon is g= 1.6 m/s² and given that the initial velocity V_y1 = 1290 m/s.

First let us find the time taken to reach one fifth of the velocity V_{y1 ,}that is V_y1 / 5 = 1290 / 5 = 258 m/s

Hence V_y = 258 m/s

hence the time taken to reach that velocity should be ,

Hence

Applying t = 645 in the first formula we get the vertical distance as

Hence the height at which the velocity would become one fifth of the initial velocity would be 499.23 km.

I hope the solution helps.... Feel free to comment and discuss further... Cheers :)

genius_generous answered 2 years ago

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