posted by veronica on .
A cannonball is shot from a cannon on a cliff at a height of 18.9 m at a ship on the ocean. The cannonball is shot with an initial velocity of 15.1 m/s at an angle of 24.7 degrees above the horizontal.
#1- What is the total maximum height of the cannonball (the height with respect to the ocean)?
#2- What is the total time in the air for the cannonball?
#3- What is the total range of the cannonball?
First, resolve the velocity into the horizontal and vertical components,
The horizontal velocity will stay constant, while the vertical velocity will be subject to the influence of gravity.
The height above the ocean at time t is given by the equation:
g=9.8 m/s², acc. due to gravity
y0=18.9 m = height of cliff
The vertical velocity at time t is given by
a. The cannonball reaches the highest point at time t when v(t)=0
b. When the cannonball hits the sea, y(t)=0. The time t1 when this happens is
Solve for t1.
c. Since the horizontal velocity is constant at vx, the total range is
Distance = vx*t1
Post if you have questions or if you wish to have your answers checked.