A small canon is placed on top of a fortification the cannon ball leaves the muzzle of the canon with a speed of 85 m/s

Speed of  cannonball just before it hits the ground is  90.77 m/s

Explanation:

Complete Question

A small cannon is placed on top of a fortification. the cannonball leaves the muzzle of the cannon with a speed of 85 m/s at an angle of 25°c above the horizontal. just before the cannonball hits the ground, the vertical component of velocity is 48 m/s downward. what is the speed of the cannonball just before it hits the ground? ignore air resistance.

Solution

Given  

Speed = 85 m/s

The angle = 25 degrees

When it will hit the ground, then vertical velocity = 48 m/s

However, in the projectile motion, the horizontal component will not change  

Vr = V cos (theta) = 85 * cos25

Speed of  cannonball just before it hits the ground is  

V’ = Sqrt (48^2 + (85 * cos 25)^2) = 90.77 m/s

resonant frequency

wavelength divides in half and speed stays the same.