Posted by **veronica** on Tuesday, November 24, 2009 at 1:25am.

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?

- physics -
**MathMate**, Tuesday, November 24, 2009 at 2:45pm
First, resolve the velocity into the horizontal and vertical components,

v0=15.1 m/s

θ=24.7°

vx=v0 cosθ

vy=v0 sinθ

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:

y(t)=y0+vy*t-(1/2)gt²

where

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

v(t)=vy-g*t

a. The cannonball reaches the highest point at time t when v(t)=0

v(t)=vy-g*t=0

tmax=vy/g

Maximum height

y(tmax)=y0+vy*tmax-(1/2)g*tmax²

b. When the cannonball hits the sea, y(t)=0. The time t1 when this happens is

y(t1)=y0+vy*t1-(1/2)gt1²=0

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.

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