Question

Two 4.0-kg blocks are tied together with a compressed spring between them. They are thrown from the ground with an initial velocity of 35 m/s, 45° above the horizontal. At the highest point of the trajectory they become untied and spring apart. About how far below the highest point is the center of mass of the two-block system 2.0 s later, before either fragment has hit the ground?

Answer 1

Answer:

The distance is 20 m.

Explanation:

Given that,

Mass of two blocks = 4.0 kg

Initial velocity = 35 m/s

Angle = 45°

Time = 2 sec

After, they become untied, center of mass trajectory will not change:

We need to calculate the distance

Using equation of motion

$s=ut-\dfrac{1}{2}gt^2$

Where, u = initial velocity

g = acceleration due to gravity

t = time

$s=0\times2-\dfrac{1}{2}\times9.8\times2^2$

$s=-19.6 = -20\ m$

$|s|=20\ m$

Hence, The distance is 20 m.