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