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 ?

2 Answers
May 13, 2018

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

enter image source here

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.

enter image source here

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^@

May 25, 2018

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