Help with symmetric and asymmetric projectiles?

Question

I have no idea of how projectiles work and i need lots help in my physics class thank you very much if you can help me.

Symmetric
object launched at 60 degrees Vx=120m/s whatis the Vi?
How high does it go and what is the impact velocity.

Asymmetric
Mortar 250meters above a target, target 750meters horizontally away.If shells burts 6meter above ground If the Vi=225m/s what is the launch angle?

How do you find the range and max height with only the Vi, launch angel, Sine, Cosine, Vx, and Vi,y?

It would be very helpful if you can show the steps. Thank you very much!

Answer 1

Symmetric:

Knowing the angle and vx, vi is related by
vi*cos(θ)=vx
IN this case
vi=120/cos(60)
vi=240 m/s

Using energy to calculate the height
.5*m*vy^2=m*g*h
simplify
h=.5*vy^2/g
vy=vi*sin(θ)
plug in the numbers
vy=207.8461

h=2202 m

The impact velocity will be the same vx, with the same magnitude vy, only negative. The resultant has magnitude equal to vi

Asymmetric:

I will assume that the mortar burst is 6 m above the target
Use the equations of motion
y(t)=250+vi*sin(θ)*t-.5*g*t^2
x(t)=vi*cos(θ)*t
when y(t)=6 and x(t)=750, the mortar bursts
solve for θ

6=250+225*sin(θ)*t-.5*g*t^2
750=225*cos(θ)*t

0=244+225*sin(θ)*t-.5*9.81*t^2
750=225*cos(θ)*t

10/(3*cos(θ))=t

0=224+2250*sin(θ)/(3*cos(θ))-
.5*9.81*100/(9*cos^2(θ))


I am sure there is some trick to finding θ in an elegant way. I don't know what it is, so I set up a spreadsheet to solve for it. There will be two values.

-13.9390 degrees
and
85.9172 degrees

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