Question

Se lanza verticalmente hacia arriba un cuerpo que para por un punto A con una rapidez de 54 m/s y por otro punto B situado mas arriba con 24 m/s.calcular :a) tiempo en ir desde A hasta B;b )la altura vertical entre dichos puntos.
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Answer 1

1) The time taken is 3.06 s

2) The distance between the two points is 119.4 m

Explanation:

1)

The motion of the object is a free fall motion, so it is a uniformly accelerated motion with constant acceleration $a=g=-9.8 m/s^2$ towards the ground. So, we can solve the problem by using the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

In this problem we have

u = 54 m/s is the initial velocity (at point A)

v = 24 m/s is the final velocity (at point B)

$a=-9.8 m/s^2$ is the acceleration (negative since it is downward)

Solving for t,

$t=\frac{v-u}{a}=\frac{24-54}{-9.8}=3.06 s$

2)

For this part of the problem, we can use another suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

In this problem we have

u = 54 m/s

v = 24 m/s

$a=-9.8 m/s^2$

Solving for s, we find the distance between point A and point B, which is the distance covered by the object:

$s=\frac{v^2-u^2}{2a}=\frac{24^2-54^2}{2(-9.8)}=119.4 m$

Learn more about accelerated motion:

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