On new year eve,a fire work rocket is launched vertically into the night sky with an initial speed of45m/s .After 2.55s ,the coasting rocket then explode and break into two pieces with an equal mass .Each pieces follow trajectory of 45 degree to the verticle . Find their speed immediately after explosion.
v = 45 - 9.81 t = 45 - 9.81(2.55) = 20 m/s
initial momentum up = 20 m
final momentum up = 20 m = (1/2)m v cos 45 + (1/2)m v cos 45
final up = initial up
m v cos 45 = 20 m
v = 20/cos45 = 20 sqrt 2
To find the speed of each piece of the rocket immediately after the explosion, we can break down the problem into two components: horizontal and vertical.
First, let's find the vertical component of the velocity for each piece of the rocket. We know that the initial speed of the rocket is 45 m/s, and it follows a vertical trajectory for 2.55 seconds. The equation to calculate the vertical displacement (Δy) is:
Δy = v0y * t + 0.5 * g * t^2
where v0y is the initial vertical velocity, t is time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the rocket is launched vertically, the initial vertical velocity (v0y) is equal to the initial speed of the rocket. Thus,
v0y = 45 m/s
Substituting the given values into the equation, we get:
Δy = (45 m/s) * (2.55 s) + 0.5 * (9.8 m/s^2) * (2.55 s)^2
Δy ≈ 45 m + 32.1 m ≈ 77.1 m
Now, let's find the horizontal component of the velocity for each piece. We know that the trajectory of each piece is at a 45-degree angle with the vertical. Therefore, the initial horizontal velocity (v0x) for each piece is equal to the initial speed of the rocket divided by √2.
v0x = 45 m/s / √2 ≈ 31.9 m/s
Now, to find the speed (v) immediately after the explosion, we can use the Pythagorean theorem:
v = √(v0x^2 + v0y^2)
Substituting the values we found:
v = √((31.9 m/s)^2 + (77.1 m/s)^2)
v ≈ √(1017.61 m^2/s^2 + 5953.41 m^2/s^2)
v ≈ √(6971.02 m^2/s^2)
v ≈ 83.5 m/s
Therefore, the speed of each piece of the rocket immediately after the explosion is approximately 83.5 m/s.