Question

Se lanza un cuerpo verticalmente hacia arriba con una velocidad de 60m/s. Calcular a) Que altura se hallara el cuerpo al cabo de 2s, b)a qué altura máxima llega? C) cuánto tiempo se demora en llegar a la altura máxima y en qué tiempo regresa nuevamente al sitio donde fue lanzado.

Answer 1

(a) 100.38 m

(b) 183.49 m

(c) 6.12 seconds and 12.24 seconds.

Step-by-step explanation:

Given that a body is thrown vertically upwards with a velocity of 60 m/s.

Assuming that upward direction is positive

So, the initial velocity, u=60 m/s

As the acceleration due to gravity, g=9.81 m/s^2, acts in the downward direction, so

a=-9.81 m/s^2

(a) By using the equation of motion, the displacement of the body after time t=2 seconds, is

`s=ut+\frac 1 2 at^2 \\\\ s=60* 2 +\frac 1 2 (-9.81)2^2 \\\\`

s=100.38 m

Hence, the height of the body will be 100.38 m after 2 seconds.

(b) At the maximum height, the velocity of the object is zero, so

v=0.

The maximum height is the displacement of the body where it attains zero velocity.

Using the equation of the motion.

`v^2=u^2+2as \\\\0= 60^2 +2(-9.81)s \\\\s=\frac {3600}{2* 9.81} \\\\`

s=183.49 m

Hence, the maximum height of the body is 183.49 m.

(c) Time to reach the maximum height

v=u+at

0=60 +(-9.81)t

t=60/9.81

t=6.12 seconds.

The body will take 6.12 seconds to reach the maximum height.

As the time to reach the maximum height from the initial position is the same as the time to reach the initial position from the maximum height.

So, the total time to returns again to the place where it was launched

= 2 x 6.12 = 12.24 seconds.