A 200 g model rocket is observed to rise 100 m above the ground after launch. What must have been the launch speed of the rocket at the ground?
I wonder if PEFinal=KEinitial
mg*100= 1/2 m vi^2
solve for vi
To determine the launch speed of the rocket, we can use the principle of conservation of energy. The potential energy gained by the rocket when it reaches a height of 100 m is equal to the kinetic energy it had when it was initially launched from the ground.
The potential energy (PE) of an object at a certain height is given by the formula PE = mgh, where m is the mass of the object (200 g = 0.2 kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
So, in this case, the potential energy gained by the rocket is PE = (0.2 kg)(9.8 m/s²)(100 m) = 196 J.
The kinetic energy (KE) of an object is given by the formula KE = (1/2)mv², where m is the mass of the object and v is its velocity.
Since the rocket has no initial potential energy when it is launched from the ground, all the initial energy is in the kinetic form. Therefore, the kinetic energy of the rocket when it was launched is equal to the potential energy gained.
Setting the potential energy and kinetic energy equal to each other, we have:
KE = PE
(1/2)mv² = mgh
Simplifying the equation, we get:
v² = 2gh
Now, we can substitute the values into the equation and solve for v:
v² = 2(9.8 m/s²)(100 m)
v² = 1960 m²/s²
v = √(1960 m²/s²)
v ≈ 44.2 m/s
Therefore, the launch speed of the rocket at the ground must have been approximately 44.2 m/s.