A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's acceleration has components a(t)_x=alpha*t^2 and a(t)_y=beta-gamma*t, where alpha = 2.50 , beta= 9.00 , and gamma= 1.40 . At the rocket is at the origin and has velocity v_0=v_0xi+v_0yj with v_0x= 1.00 m/s and v_0y= 7.00 m/s. What is the maximum height reached by the rocket
To find the maximum height reached by the rocket, we need to determine the time at which the rocket reaches its highest point. At the highest point, the vertical component of its velocity will be zero.
Let's start by finding the time at which the rocket reaches its highest point. We know that the vertical component of the acceleration is given by a(t)_y = beta - gamma*t, where beta = 9.00 and gamma = 1.40.
Setting a(t)_y = 0, we can solve for t:
beta - gamma*t = 0
9.00 - 1.40*t = 0
t = 9.00 / 1.40
t ≈ 6.43 seconds
So, the rocket reaches its highest point at approximately 6.43 seconds.
Now, let's find the maximum height reached by the rocket. To do this, we need to calculate the vertical displacement at time t = 6.43 seconds. We can use the equation of motion for displacement in the y-direction:
y(t) = v_0y * t + (1/2) * a(t)_y * t^2
Substituting the given values:
y(6.43) = (7.00 m/s) * (6.43 s) + (1/2) * (9.00 - 1.40 * 6.43) * (6.43 s)^2
Evaluating the above expression will give us the maximum height reached by the rocket.