Respuesta :

skyluke89
The kinetic energy of the object just after it starts its motion from the Earth surface is
[tex]K= \frac{1}{2}mv^2 [/tex]
where m is the object mass and [tex]v=3.7 km/s=3700 m/s[/tex] its initial speed.
When it reaches its maximum height, the object speed is zero and all its kinetic energy converted into gravitational potential energy, which is
[tex]U=mgh[/tex]
where [tex]g=9.81 m/s^2[/tex] and h is the maximum height reached by the object.

Since the energy of the object must be conserved, K=U, therefore we can write
[tex] \frac{1}{2}mv^2 =mgh[/tex]
and we can solve to find h, the maximum height:
[tex]h= \frac{v^2}{2g}= \frac{(3700 m/s)^2}{2\cdot 9.81 m/s^2}= 6.98 \cdot 10^5 m=698 km[/tex]

The maximum height it reaches. will be 698 km.it is the height achieved by the body when a body is thrown at the same angle.

What is the maximum height achieved in projectile motion?

it is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion. the maximum height of motion is given by

The given data in the problem is;

u is the initial speed of 3.7 km/s.=3700 m/sec²

h is the maximum height=?

g is the gravitational acceleration = 9.81 m/sec

The maximum height is obtained as;

[tex]\rm h= \frac{v^2}{2g} \\\\ \rm h= \frac{(3700)^2}{2\times 9.81 } \\\\ \rm h=698 Km[/tex]

Hence the maximum height it reaches. will be 698 km.

To learn more about the projectile motion refer to the link;

https://brainly.com/question/6261898

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