A rocket is launched from the origin with an acceleration of 20.0 m/s2 in a straight line at an angle of 30.0 degrees above the horizontal. The launch acceleration lasts for 2.00 seconds at which time the fuel is exhausted. The rocket then falls with an acceleration of 9.80 m/s2 downward. What is the time it takes to reach maximum height?

4.04 seconds

Explanation:

In order to solve this problem, we must keep in mind that the initial 2.00 seconds at which the fuel is exhausted does not signal when the rocket reaches its maximum height.

From this moment on, the rocket has a downward acceleration of 9.8 m/s² but it still has an upwards velocity, which we will calculate. This upwards velocity keeps the rocket moving up for a certain period of time, which we will also calculate.

For this problem, let's set the upwards direction to be positive and the downwards direction to be negative.

To find the time that the rocket takes to reach its maximum height, we are going to use the initial 2.00 seconds and find the additional seconds it takes to reach a final vertical velocity of 0 m/s. This represents the time at which the rocket stops moving and heads in the downwards direction, which takes place after the rocket has reached its maximum height.

The time in the air of an object in projectile motion can be found using this equation, derived from one of the constant acceleration kinematic equations.

Time in the air (projectile motion):

Solve for by plugging in 2.00 seconds for t, 30 degrees for theta, and 20.0 m/s² for acceleration, since this is not a constant acceleration problem.

Multiply sin(30) and 2 together.

Multiply 20 to both sides of the equation.

Now we know that the initial velocity of the rocket is 40 m/s. We need to solve for the vertical component of the rocket's velocity in order to solve for the additional time it took after the 2.00 seconds to reach its maximum height.

Vertical component:

The vertical component of the velocity vector is 20 m/s.

Now, in order to solve for the additional seconds that the rocket took to reach its maximum height, let's use one of the kinematic constant acceleration equations that uses the variables , , , and .

Since we are trying to solve for time, we need to use this equation in terms of the vertical direction, aka the y-direction. Time is the same in either case.

The final vertical velocity of this rocket is 0 m/s at the top, or its maximum height. We found that the vertical component, aka the rocket's initial vertical velocity, is 20 m/s. The acceleration is given to us: -9.8 m/s² (since it's falling downwards, the acceleration must be negative because we already established this in the beginning).

We are trying to solve for time t. Substitute the known values into the equation.

Subtract 20 from both sides of the equation.

Divide both sides of the equation by -9.8.

Now we can take this time and add it to the initial 2.00 seconds of the rocket. This time we just solved for is the time after these initial seconds that the rocket kept going upwards, since its initial vertical velocity was not 0 m/s yet.

The time that the rocket takes to reach its maximum height is 4.04 seconds.

4.04 second is your answer hope it help you