pereus said:Homework Statement
The speed of a projectile when it reaches its maximum height is 1/4 the speed when the projectile is at 1/4 its maximum height.
What is the initial projection angle?
Answer in units of ◦
Homework Equations
v2 = v02 + 2a(delta s)
4Vx = v1/4
The Attempt at a Solution
I've tried substituting v1/4 for 4vx and some other things, but I lost those papers.
pereus said:I know that the velocity at the top is Vx. That is how I was able to make 4vx = to 4v1/4
pereus said:At the top the projectile is moving with the velocity Vx which is equal to v0cos(theta). There is no velocity in the y direction
A projectile at 1/4 height refers to an object that is launched or thrown into the air at an angle and reaches a height that is 1/4 of its maximum height before falling back to the ground.
The height of a projectile at 1/4 height can be calculated using the formula h = (1/4) * (v^2/g), where h is the height, v is the initial velocity, and g is the acceleration due to gravity.
The height of a projectile at 1/4 height is affected by the initial velocity, launch angle, and the acceleration due to gravity.
Air resistance can reduce the height of a projectile at 1/4 height by slowing down its speed and reducing its maximum height. This is because air resistance creates a force that acts against the motion of the projectile.
The height of a projectile at 1/4 height can be increased by increasing the initial velocity, increasing the launch angle, or reducing the effects of air resistance. This can be achieved by using a more aerodynamic projectile or launching it in an environment with less air resistance, such as in a vacuum.