A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point.What is the angle of launch?

What is the launch speed?

The answers is 69 degrees. I don't know why

To find the angle of launch, we can use the following trigonometric equation:

tan(θ) = (H / L)

where θ is the angle of launch, H is the height reached by the rocket, and L is the horizontal distance traveled by the rocket.

In this case, H = 72.3 m and L = 111 m.

Plugging these values into the equation, we can solve for θ:

tan(θ) = (72.3 / 111)

Taking the arctan of both sides, we can find θ:

θ = arctan(72.3 / 111)

Using a calculator, we find θ ≈ 33.8 degrees.

To find the launch speed, we can use the following kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity (which is 0 m/s at the highest point of the rocket's trajectory), u is the initial velocity (the launch speed we want to find), a is the acceleration (equal to -9.8 m/s^2 due to gravity), and s is the vertical distance traveled by the rocket.

In this case, s = 72.3 m.

Rearranging the equation, we can solve for u:

u^2 = -2as

u = √(2as)

Plugging in the known values, we get:

u = √(2 * (-9.8) * 72.3)

Using a calculator, we find u ≈ 29.6 m/s.

Therefore, the angle of launch is approximately 33.8 degrees and the launch speed is approximately 29.6 m/s.

To find the angle of launch and the launch speed, we can use the principles of projectile motion and trigonometry.

First, let's consider the vertical motion of the rocket. We know that the maximum height reached is 72.3 m. At the peak of its trajectory, the vertical velocity of the rocket becomes 0, and the only force acting on it is gravity, causing it to fall back down. We can use this information to calculate the time it takes for the rocket to reach its peak.

We can use the kinematic equation:
v_y = u_y + a*t,

where v_y is the final vertical velocity (0 m/s at the peak), u_y is the initial vertical velocity, a is the acceleration due to gravity (approximately -9.8 m/s^2), and t is the time.

Since the rocket only has an initial vertical velocity, we can write:
0 = u_y - 9.8*t.

Simplifying the equation:
u_y = 9.8*t.

Next, we can consider the horizontal motion of the rocket. We know that the rocket lands 111 m from the launch point. The horizontal velocity is constant throughout the motion.

We can use the kinematic equation:
s_x = u_x*t,

where s_x is the horizontal distance traveled, u_x is the initial horizontal velocity, and t is the time of flight.

Since the horizontal velocity remains constant, we can write:
s_x = u_x*t.

Now, let's relate the vertical and horizontal motions using the launch angle. The launch velocity (V) can be split into vertical (V_y) and horizontal (V_x) components. We can use trigonometry to find these components:

V_y = V*sin(θ),
V_x = V*cos(θ),

where θ is the launch angle.

From the initial vertical velocity equation:
u_y = V_y = 9.8*t.

From the horizontal distance equation:
s_x = V_x*t = V*cos(θ)*t.

Now, we have two equations involving time and the launch angle. To eliminate time, we can solve for it in one equation and substitute it into the other equation.

From the horizontal distance equation:
t = s_x / (V*cos(θ)) = 111 / (V*cos(θ)).

Substituting this expression for time into the vertical velocity equation:
u_y = 9.8 * (111 / (V*cos(θ))).

Now, we can solve this equation for the launch angle (θ). Rearranging the equation:
tan(θ) = (u_y * V) / (9.8 * s_x).

Substituting the known values:
tan(θ) = (72.3 * V) / (9.8 * 111).

Finally, we can use inverse tangent (arctan) to find the launch angle:
θ = arctan((72.3 * V) / (9.8 * 111)).

As for the launch speed (V), we need to know either the launch angle or the initial velocity (V) to find it. Without more information, it is not possible to determine the launch speed without making assumptions.

I don't think it can be worked, without knowing how long the engine burns. The rocket is not a projectile until the fuel burns out. The launch speed is easy, it starts at zero veloicty, the rocket burns and it is accelerated upward.