wowdusk said:Homework Statement
A projectile is launched with a speed of 40 m/s at an angle of 60 degrees above the horizontal. Use conservation of energy to find the maximum height reached by the projectile during its flight.
Homework Equations
KEi+PEi=KEf+PEfThe Attempt at a Solution
i am not sure where vf would come from. Would it be just Vi*cos(60)?
I am not sure why that makes sense.
Does Vix=Vf?...and why?
wowdusk said:why do i need the vertical component of Vi
I don't know how to find the V at the heighest point...
At the heighest point is the V in vertical direction 0 anyway?
Projectile motion is the motion of an object that is launched or thrown into the air, and is affected by both gravity and the initial velocity of the object.
The maximum height of a projectile is determined by the initial velocity of the object, the angle at which it is launched, and the force of gravity acting on the object. These factors can be calculated using mathematical equations to determine the maximum height of the projectile.
The formula for calculating the maximum height of a projectile is h = (v2sin2θ) / (2g), where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.
Yes, the maximum height of a projectile can be greater than the initial height if the launch angle is greater than 90 degrees. In this case, the object would follow a parabolic path and reach its maximum height before falling back to the ground.
The maximum height of a projectile can be affected by various factors such as the initial velocity, launch angle, air resistance, and the force of gravity. Other external factors such as wind and air density can also affect the maximum height of a projectile.