A 5 kg model rocket is lifted off the ground by a force of 80 N. At a height of 10 m, what is the speed of the rocket? Use g = 10 m/s2Question 1Answera.17.89 m/sb.14.76 m/sc.21.99 m/sd.10.95 m/se.35.9 m/s

To solve this problem, we need to use the principle of conservation of energy. The total energy of the rocket at the start (when it's on the ground) is equal to the total energy of the rocket when it's at a height of 10 m.

1. First, calculate the work done by the force to lift the rocket off the ground. Work done (W) is equal to force (F) times distance (d). In this case, the force is 80 N and the distance is 10 m. So, W = F * d = 80 N * 10 m = 800 J (Joules).

2. This work done becomes the kinetic energy of the rocket when it's at a height of 10 m. So, the total energy of the rocket at this point is its kinetic energy (KE) plus its potential energy (PE). The potential energy is given by m * g * h, where m is the mass of the rocket (5 kg), g is the acceleration due to gravity (10 m/s^2), and h is the height (10 m). So, PE = 5 kg * 10 m/s^2 * 10 m = 500 J.

3. Therefore, the kinetic energy of the rocket at this height is the total energy minus the potential energy, which is 800 J - 500 J = 300 J.

4. The kinetic energy is also given by 1/2 * m * v^2, where m is the mass of the rocket and v is its speed. We can set this equal to the kinetic energy we found and solve for v: 300 J = 1/2 * 5 kg * v^2.

5. Solving for v gives v = sqrt((2 * 300 J) / 5 kg) = sqrt(120 m^2/s^2) = 10.95 m/s.

So, the speed of the rocket at a height of 10 m is 10.95 m/s. The correct answer is (d) 10.95 m/s.