Question

Marvin the Martian needs to get back home. Marvin is 321,770 m from his home on Mars. He decides the quickest way to get home is to use his canon to fire himself into flight. He aims the the canon at an angle of 25 degrees. When the canon is fired Marvin the Martian is launched into flight at an initial velocity of 1250 m/s. The question is will his plan work

Answer 1

Answer:

y = 14238 m.    the height of the rocket is much less than this distance therefore the plan will not work.

Explanation:

Let's analyze this exercise, so that the Martian's plan works, the vertical height of the body must be zero when it is more than half of the way to the planet Mars, this is so that Mars attracts it and can arrive.

Let's calculate the maximum height of the launch

          $v_{y} ^2 = v_{oy}^2 - 2 g y$

at the highest point $v_{y}$ = 0

          y = v_{oy}² / 2g

          y = (v₀ sin θ)² / 2g

let's calculate

          y = (1250 sin 25)² /2 9.8

          y = 14238 m

In the exercise, indicate that the distance to Mars is h = 321770 m, half of this distance is

          h / 2 = 160885 m

therefore the height of the rocket is much less than this distance therefore the plan will not work.

The height reached is low, so it is not necessary to take into account the variation of g with height