Question

A shell (a large bullet) is shot with an initial speed of 20 m/s, 60 degrees above the horizontal. At the top of the trajectory, the bullet explodes into two fragments of equal mass. One fragment has a speed of zero just after the explosion and simply drops straight down. How far from the gun does the other fragment land, assuming that the ground is level and that the air drag is negligible.

Answer 1

The horizontal displacement of the other fragment is equal to the horizontal displacement of the shell before the explosion.

Step-by-step explanation:

To find the horizontal displacement of the shell when it explodes, we can analyze the projectile motion of the shell before the explosion and the subsequent vertical motion of the fragments after the explosion.

Before the explosion, the shell follows a parabolic trajectory which can be divided into horizontal and vertical components. The horizontal displacement can be found using the formula:

Horizontal displacement = initial horizontal velocity * time

After the explosion, one fragment drops straight down, so its horizontal displacement is zero. The other fragment continues to move horizontally with the same initial horizontal velocity as the shell and lands at a distance equal to the initial horizontal displacement of the shell.

Therefore, the horizontal displacement of the other fragment, assuming negligible air drag, is equal to the horizontal displacement of the shell before the explosion.

Answer 2

17.656 m

Step-by-step explanation:

Initial speed u = 20 m/s

angle of projection α = 60°

at the top of the trajectory, one fragment has a speed of zero and drops to the ground.

we should note that the top of the trajectory will coincide with halfway the horizontal range of the the projectile travel. This is because the projectile follows an upward arc up till it reaches its maximum height from the ground, before descending down by following a similar arc downwards.

To find the range of the projectile, we use the equation

R =
`(u^(2)sin2\alpha )/(g)`

where g = acceleration due to gravity = 9.81 m/s^2

Sin 2α = 2 x (sin α) x (cos α)

when α = 60°,

Sin 2α = 2 x sin 60° x cos 60° = 2 x 0.866 x 0.5

Sin 2α = 0.866

therefore,

R =
`(20^(2)*0.866 )/(9.81)` = 35.31 m

since the other fraction with zero velocity drops a top of trajectory, distance between the two fragments assuming level ground and zero air drag, will be 35.31/2 = 17.656 m