The launching speed of a certain projectile is 6.4 times the speed it has at its maximum height. Calculate the elevation angle at launching.
I have noo idea how to start this one. Please help!
The vertical component of velocity at max height is zero. The horizontal component remains Vh during flight, and that is the speed at max height. Let Voy be the initial vertical component of the launch speed.
(launch speed)^2 = Voy^2 + Vh^2
= (6.4 Vh)^2 = 40.96 Vh^2
Voy^2 = 39.96 Vh^2
Voy = 6.32 Vh
launch angle arctan(Voy/Vh) = arctan 6.32 = 81.0 degrees
How did you come up with 39.96 Vh^2?
I squared 6.4 and subracted 1. See the formula
Voy^2 + Vh^2 = 40.96 Vh^2
This is probably really obvious, but I'm still not undertanding why you have to subtract one.
To solve this problem, we can use the equations of projectile motion. The key concept we will be using is that the speed of the projectile when it reaches its maximum height is zero.
Let's assume the speed of the projectile at its maximum height is V_max.
Given that the launching speed of the projectile is 6.4 times the speed at its maximum height, we can express it as:
V_launch = 6.4 * V_max
We know that at the maximum height, the vertical component of the velocity is zero. Therefore, we can write the equation for the vertical component of velocity at launch as:
V_launch_vertical = V_launch * sin(theta) = 0
Where theta is the elevation angle at launching.
Now, we need to find the value of theta that satisfies this equation.
Since V_launch is non-zero, we can divide both sides of the equation by V_launch to get:
sin(theta) = 0 / V_launch
Since any number divided by zero is undefined, this equation tells us that there is no unique solution for theta.
Therefore, the elevation angle at launch cannot be determined based on the given information.