Question

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?

Answer 1

`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)`

Answer 2

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.

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"`