In a projectile motion, if we measure the maximum
height h and horizontal range R
(1) How to get launch angle θ in terms of h and R
(2) 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?
Let θ be the launch angle. Assume the ground is level. Let V be the launch speed.
R = 2 (V^2/g) sin θ cos θ
h = V^2 sin^2 θ / (2g)
R = 4 h cosθ/sin θ = 4 h cotθ
Check my thinkng.
21
To find the launch angle and launch speed in a projectile motion, we can use the equations of motion and some trigonometry. Here's how you can get the answers:
(1) To find the launch angle in terms of the maximum height h and horizontal range R, we can use the following trigonometric equation:
tan(θ) = h / R
Solving this equation for θ, we can take the inverse tangent (arctan) of both sides:
θ = arctan(h / R)
Therefore, the launch angle θ is given by the arctan of h divided by R.
(2) Given that the maximum height is h = 72.3 m and the horizontal range is R = 111 m, we can now substitute these values into the equation we derived in step (1):
θ = arctan(72.3 / 111)
Using a calculator, we find that the launch angle θ is approximately 34.45 degrees.
To find the launch speed, we can use the equations of motion for projectile motion. The horizontal range R is related to the launch speed v and the launch angle θ by the formula:
R = (v^2 * sin(2θ)) / g
Here, g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging this equation, we can solve for v:
v = sqrt((R * g) / sin(2θ))
Now, we can substitute the values R = 111 m, θ = 34.45 degrees, and g = 9.8 m/s^2 into the equation:
v = sqrt((111 * 9.8) / sin(2 * 34.45))
Using a calculator, we find that the launch speed v is approximately 40.75 m/s.
Therefore, the angle of launch is approximately 34.45 degrees, and the launch speed is approximately 40.75 m/s.