Aprojectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizontal. at some time after its launch, an explosion splits the projectile into two pieces. one piece, of mass 6.5 kg, is observed at 1.42 seconds after the launch to be at a height of 5.9 m and at a horizontal distance of 13.6 m from the launch point. find the location of the second fragment at that same instant of time.

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle,

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range,

Time of flight, t' =

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

Distance traveled vertically:

Now,

x' = 3.535 m

Similarly,

The location of the other fragment is at (3.535, 1.162)

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s