A ball is launched as a projectile?

Question

A ball is launched as a projectile with initial speed v at an angle (theta) above the horizontal. Using conservation of energy, find the maximum height hmax of the ball's flight.
Express your answer in terms of v,g , and theta.

Answer 1

It's a good thing the problem said to use conservation of energy, because that's the easiest approach.

We can make this problem a bit easier. Rather than to look at the problem as a ball moving in the x and y directions, look at it as if it were only in the y direction. After all, that's where the hmax comes from!

To do this, we need to determine the speed the ball was shot in the y direction. Draw a right triangle. v is the hypotenuse, vy is the vertical leg and vx is the horizontal leg. θ is the base angle. So vy = v sin θ. This is the speed we will be working with in our calculations.

Initially the ball has PE = 0 since h = 0. Finally the ball has KE = 0 since vy = 0.


Ei = Ef
½m (v sin θ)² = mgh

m's cancel, as usual

½ (v sin θ)² = gh

h = (v sin θ)² / (2g)


Done!

Answer 2

before using conservation of energy you need to find velocity at highest point
at highest point its velocity would be equal to initial velocity along x axis=vcost
t is theta

1/2mv^2=1/2m(vcost)^2+mgh
1/2mv^2(1-cos^2t)=mgh
1/2mv^2(sin^2t)=mgh

h=(vsint)^2 / 2g

Answer 3

not a good question you know.