A cannon shoots a projectile at 24 degrees from the horizontal. It lands on level ground 3000m down range. a) Find the initial velocity b) Calculate its maximum height?

2 Answers
Feb 20, 2017

#sf((a))#

#sf(199color(white)(x)"m/s")#

#sf((b))#

#sf(333.9color(white)(x)m)#

Explanation:

Treat the vertical and horizontal components separately and then eliminate the time of flight.

#sf((a))#

The horizontal component of the velocity is constant so we can write:

#sf(vcostheta=d/t)#

#:.##sf(t=d/(vcostheta)" "color(red)((1)))#

#sf(t)# is the time of flight

#sf(d)# is the horizontal range

For the vertical component we can use the general equation:

#sf(v=u+at)#

This becomes:

#sf(0=vsintheta-("g"t)/2)#

Note we divide the total time of flight by 2 because the time to reach maximum height (what we want) is the total time of flight / 2.

#:.##sf(vsintheta=("g"t)/2" "color(red)((2)))#

Now we can substitute the value of #sf(t)# from #sf(color(red)((1))# into #sf(color(red)((2))rArr)#

#sf(vsintheta=(gd)/(2vcostheta))#

#:.##sf(gd=2v^2(sinthetacostheta))#

We can simplify this using the trig identity:

#sf(sin2theta=2sinthetacostheta)#

#:.##sf(gd=v^2sin2theta)#

This gives #sf(d=(v^2sin2theta)/(g))# and is a useful formula which you can use to get the range.

However, we need to get v :

#sf(v^2=(dg)/(sin2theta))#

Putting in the numbers:

#sf(v^2=(9.81xx3000)/(sin48)=(9.81xx3000)/(0.7431)=39604.36)#

#:.##sf(v=sqrt(39604.36)=199color(white)(x)"m/s")#

#sf((b))#

We can use:

#sf(v^2=u^2+2as)#

This becomes:

#sf(0=(vsintheta)^2-2gh)#

#:.##sf(h=(vsintheta)^2/(2g))#

#:.##sf(h=(199xx0.4067)^2/(2xx9.81)=333.9color(white)(x)m)#

Feb 20, 2017

a) The initial velocity is 199 m/s; b) The maximum height is 334 m.

Explanation:

The general path of a projectile is shown below.

keisan.casio.com

We can derive an equation for the initial velocity #v# of the cannonball given the range #"R"# and the angle of elevation #θ#.

For motion in the horizontal direction,

#v_x = vcosθ#

#v_x# is constant, so at any time the #x# position of the cannonball is

#x = v_xt = vtcosθ#

For motion in the vertical direction,

#v_y = vsinθ#

#v_y# changes with time, because the cannonball is acted on by gravity. Thus,

#v_y = vsinθ - g·t#

At maximum height,

#v_y = 0#

#0 = vsinθ - g·t#

The time to reach the top of the parabola is

#t_"top" = (vsinθ)/g#

It will take the same time for the cannonball to fall, so the total time in the air is

#t = (2vsinθ)/g#

Knowing the time allows us to find the range of the cannon:

#R = v_xt = vcosθ × (2vsinθ)/g = (2v^2cosθsinθ)/g = (v^2sin2θ)/g#

Rearranging gives

#v = sqrt((gR)/(sin2θ))#

#g = "9.81 m·s"^"-2"#
#R = "3000 m"#
#θ = 24°#

#v = sqrt(("9.81 m·s"^"-2" × "3000 m")/sin(48°)) = sqrt("39 600 m"^2"s"^"-2") = "199 m·s"^"-1"#

b) Maximum height

#t_"top" = (vsinθ)/g#

The distance traveled by a falling object is

#h = 1/2g·t^2#

Therefore, the maximum height is

#h = 1/2g((vsinθ)/g)^2 = (v^2sin^2θ)/(2g) = ((199 "m"·color(red)(cancel(color(black)("s"^"-1"))))^2 × sin^2(24°))/(2 × 9.81 "m"·color(red)(cancel(color(black)("s"^"-2"))))#

#h = 6551/19.62color(white)(l) "m" = "334 m"#