Um corpo de massa m= 2,0Kg é lançado horizontalmente, de uma altura h= 125m, com velocidade de módulo Vo =10m/s, como mostra a figura. Desprezando a resistência do ar e adotando g= 10m/s2 , determine: a) A energia mecânica total do corpo; b) A energia cinética do corpo a meia altura em relação ao solo; c) O tempo gasto até que o corpo atinja o solo; d) O alcance do movimento.

A) E = 2550 J

B) K = 1325 J

C) t = 5,05 s

Explanation:

A) The total mechanical energy is given by the sum of the gravitational potential energy and the kinetic energy of the body:

 (1)

m: mass of the body = 2,0 kg

g: gravitational acceleration = 9,8 m/s^2

h: height = 125 m

v: initial velocity of the body = 10 m/s

You replace the values of all variables h, m, g and v in the equation (1):

the total mechanical energy is 2550 J

B) The kinetic  energy of the corp, when it is at a height of h/2 is given by:

where

The x component of the velocity is constant in the complete trajectory, which is the initial velocity, that is, vo = vx

The y component is given by:

voy: vertical initial velocity = 0m/s

y: height = h/2 = 125/2 = 62.5 m

Then, you can calculate the velocity of the body and next, you can calculate the kinetic energy:

C) The time that body takes in all its trajectory is:

the inner core.

good luck: )

klicker

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.