A projectile is fired from ground level at an angle of 40.0° above horizontal at a speed of 30.0 m/s. What is the speed of the projectile when it has reached a height equal to 50.0% of its maximum height?

26.74 m/s

Explanation:

The vertical and horizontal of the velocity of the projectile when it first fired from ground level at an angle of 40 degrees

As the ball travels, vertically speaking its kinetic energy is converted into potential energy. When it reaches its maximum height the kinetic energy would be 0. Also let potential energy be 0 at around, we have the following energy conservation equation:

Let g = 9.81 m/s2. We can divide both sides by m

So half of the maximum height is h = H/2 = 9.48 m

Using the same energy equation at h = 9.48 m then we have

Again, divide both sides by m, and multiply by 2

If we ignore air resistance, then nothing could affect the horizontal velocity, it would stay the same at h = 9.48 m, 23m/s

massxgravity x heght=weightxheight= pe

4900=490xh

h=10