Question

A projectile is fired with initial speed v0 = 37.0 m/s from level ground at a target that is on the ground, at distance R = 13.0 m, as shown in the figure. What are the (a) least and (b) greatest launch angles that will allow the projectile to hit the ?

Answer 1

The greatest possible angle is `87.327^@`, and the least possible angle is `2.673^@`

Explanation:

Draw a diagram, and resolve the velocity into horizontal and vertical components. For now, let the angle the projectile is fired at be `theta`

Begin by finding the time the projectile is in the air for. Do this by considering the vertical component of velocity:

`s=0` total displacement when it falls is 0.
`u=37sintheta`
`a=-g`
`t=T`

Using `s=ut+1/2at^2`

`0=37Tsintheta-1/2gT^2`
`0=T(37sintheta-1/2gT)`
`T=0` or `T=(37sintheta)/(1/2g)`

`T=(74sintheta)/g`

Now we have the time, consider horizontal velocity.

`s=13`
`u=37costheta`
`a=0`
`t=(74sintheta)/g`

Use this to make an equation to find `theta`

Using `s=ut+1/2at^2`
`13=37costheta*74sintheta/g+0`
`13=2738/gsinthetacostheta`
`sinthetacostheta=(13g)/2738`

I'm unsure how to (or if you can) solve this from here. I instead plotted the graphs of `y=sinxcosx` and, (taking `g=9.81`), `y=(13*9.81)/2738`, and looked at the interception points.

From the graph, this holds if the angle is `2.673^@` or `87.327^@`.
This makes the greatest possible angle `87.327^@`, and the least `2.673^@`

Answer 2

Let the launch angle be `=theta`
Time `t` to reach the target can be found from horizontal component of velocity. Neglecting air resistance we get

`R=v_(0x)t`
`=>13=(37costheta)t`
`=>t=13/(37costheta)` ......(1)

Time of flight can also be found with the help of kinematic expression using vertical component of velocity.

`s=ut+1/2at^2`

Noting that gravity acts in a direction opposite to the direction of projection we get

`h=v_(0y)t+1/2(-g)t^2`
`0=(37sintheta)t-1/2g t^2`

Solving for `t` we get

`0=t(37sintheta-1/2g t)`
`=>t=0` and `t=(2xx37sintheta)/(g)`

Ignoring `t=0` as trivial solution as being time of projection, we get

`t=(74sintheta)/g` .......(2)

Equating (1) and (2)

`13/(37costheta)=(74sintheta)/g`
`=>2sinthetacostheta=(13g)/1369`
`=>sin 2theta=(13g)/1368` ........(3)
`=>2theta=sin^-1((13xx9.81)/1369)`
`=>2theta=5.35^@`
`=>theta=2.7^@` ......(4)

We know that for an angle `alpha`,

`sinalpha=sin(180-alpha)`

Using this identity we can write (3) as

`sin (180-2theta)=(13g)/1368`

We get

`180-2theta=5.35^@`
`=>theta=(180-5.35)/2`
`=>theta=87.3^@` .......(5)

(a) `2.7^@`
(b) `87.3^@`

.-.-.-.-.-.-.-.-.-.-

In case you can recall it is perfectly fine to use the formula for horizontal range:

`R=(2v_0^2sinθcosθ)/g`
or
`R=(v_0^2sin2θ)/g`