A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.30 m/s^2. At 10.0 s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.
How high above the launch pad will the rocket eventually go?
got it....142 m
Find the rocket's velocity at its highest point.
got it....0 m/s
Find the magnitude of the rocket's acceleration at its highest point.
got it....9.8 m/s^2
Find the direction of the rocket's acceleration at its highest point.
got it....downward
How long after it was launched will the rocket fall back to the launch pad?
got it....17.7 s
How fast will it be moving when it does so?
AHHHH!!!! I don't know....
At the top, it has no KEnergy, only PEnergy. So at the bottom, the PEnergy is converted to Kenergy.
mgh= 1/2 mv^2.
The other way..
v= sqrt(2gh) from kinetics.
Well, well, well, look who's back with another question! You've been on a roll today, my friend. Now, let's calculate how fast our falling rocket will splat back onto the launch pad.
We can use the good ol' equation v = √(2gh) to find the velocity. Please note that "h" here is the height above the launch pad.
Since our rocket started at rest, we know the initial velocity is zero. The acceleration due to gravity, "g," is approximately 9.8 m/s^2. And as we already established, the rocket's highest point is 142 meters.
Plug in the values and let's crunch those numbers! Let's see if our rocket will gently glide like a feather or come crashing down like a clumsy clown in stilts.
v = √(2 * 9.8 * 142)
v ≈ √(2788.8)
v ≈ 52.79 m/s
Ta-da! The rocket will come crashing back to the launch pad at a whopping speed of approximately 52.79 meters per second. I hope you have your safety net ready, my friend!
To find the velocity of the rocket when it falls back to the launch pad, we can use the equation relating potential energy and kinetic energy: mgh = 1/2 mv^2, where m is the mass of the rocket, g is the acceleration due to gravity, h is the height, and v is the velocity.
Since we know the height of the rocket above the launch pad is 142 m, and the acceleration due to gravity is approximately 9.8 m/s^2, we can substitute these values into the equation and solve for v.
mgh = 1/2 mv^2
m * 9.8 * 142 = 1/2 m * v^2
v^2 = 2 * 9.8 * 142
v^2 = 2786.4
v ≈ 52.8 m/s
Therefore, the rocket will be moving at approximately 52.8 m/s when it falls back to the launch pad.
To find the speed at which the rocket will be moving when it falls back to the launch pad, we can use the principle of conservation of energy. At the highest point, the rocket has all potential energy (PE) and no kinetic energy (KE). When it falls back to the launch pad, the PE is converted into KE.
We can use the equation: PE = KE
The potential energy (PE) at the highest point is given by: PE = mgh, where m is the mass of the rocket, g is the acceleration due to gravity, and h is the maximum height reached by the rocket.
The kinetic energy (KE) at the lowest point (i.e., when it falls back to the launch pad) is given by: KE = (1/2)mv^2, where v is the velocity of the rocket at that point.
Setting PE equal to KE, we have:
mgh = (1/2)mv^2
Canceling out mass (m) on both sides, we have:
gh = (1/2)v^2
Solving for v, we get:
v = sqrt(2gh)
Using the value of g as approximately 9.8 m/s^2 (acceleration due to gravity) and the maximum height you mentioned (142 m), we can substitute these values into the equation to find the velocity:
v = sqrt(2 * 9.8 m/s^2 * 142 m)
v = sqrt(2776.8)
v ≈ 52.67 m/s
Therefore, the rocket will be moving at approximately 52.67 m/s when it falls back to the launch pad.