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?
the mass of the rocket does not matter.
Given a liftoff speed of v, the height
h(t) = vt - 4.9t^2
The max height is reached when t = v/9.8, so we have
v(v/9.8) - 4.9(v/9.8)^2 = 100
v = 44.27 m/s
Or, you could just use the formula for max height
h = v^2/2g and get
v^2/19.6 = 100
v = 44.27 m/s
THANKS
To find the launch speed of the rocket at the ground, we can use the principle of conservation of energy. The potential energy gained by the rocket can be equated to the initial kinetic energy.
The potential energy gained by the rocket is given by:
Potential energy = m * g * h
Where:
m = mass of the rocket (200 g = 0.2 kg)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height gained by the rocket (100 m)
Potential energy = 0.2 kg * 9.8 m/s^2 * 100 m
Now, since the initial kinetic energy is equal to the potential energy gained, we can equate the two:
Potential energy = Initial kinetic energy
0.2 kg * 9.8 m/s^2 * 100 m = 0.5 * m * v^2
Where:
v = launch speed of the rocket at the ground
Now, we can solve for v:
0.2 kg * 9.8 m/s^2 * 100 m = 0.5 * 0.2 kg * v^2
Simplifying the equation, we can cancel out 0.2 kg, and multiply the other terms:
9.8 m/s^2 * 100 m = 0.1 * v^2
980 m^2/s^2 = 0.1 * v^2
Dividing both sides of the equation by 0.1:
9800 m^2/s^2 = v^2
Taking the square root of both sides of the equation:
v = √(9800 m^2/s^2)
Calculating the square root:
v ≈ 99 m/s
Therefore, the launch speed of the rocket at the ground must have been approximately 99 m/s.