Answer:
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
