Draw a diagram, and resolve the velocity into horizontal and vertical components. Because that now, allow the edge the projectile is fired in ~ be #theta# Begin by finding the time the projectile is in the air for. Execute this by considering the vertical component the velocity:

#s=0# full displacement once it drops 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, take into consideration 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 exactly how to (or if girlfriend can) solve this from here. I rather plotted the graphs of #y=sinxcosx# and, (taking #g=9.81#), #y=(13*9.81)/2738#, and looked in ~ the interception points. From the graph, this holds if the angle is #2.673^

You are watching: Find the angle θ above the horizontal at which the projectile should be fired.

# or #87.327^
#.This provides the greatest possible angle #87.327^
#, and the least #2.673^
# A08
may 25, 2018

Let the launch angle be #=theta#Time #t# to reach the target deserve to be discovered from horizontal ingredient of velocity. Neglecting air resistance us get

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

Time of flight can additionally be found with the assist of kinematic expression using vertical component of velocity.

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

Noting that gravity plot in a direction opposite come 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# us get

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

Ignoring #t=0# together trivial systems as gift time of projection, us get

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

Equating (1) and also (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 identification we can write (3) together

#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 usage the formula because that horizontal range:

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