How do you do this physics problem?
"A mountain climber is stranded on a ledge 30m above the ground. Rescuers on the ground want to shoot a projectile to him with a rope attached to it. If the projectile is directed upward at an initial angle of 55 degrees from a horizontal distance of 50 meters, determine the initial speed the projectile must have in order to land on the ledge."
My teacher hasn't really gone over problems like this, and I've been trying to work through it myself. It's been at least an hour for this one question. I'm desperate, help?
Comments
X = vo cos 55 t
You want vertical velocity to be zero at y = 30 meters
and x to be 50 when when vertical velocity is zero
vertical velocity =
Vy = vo sin55 - 9.8t
0 = v0 sin55 -9.8t
t = vo sin55/9.8
subitiute into x equation and solve for vo
X = vo cos 55 t
50 = vo cos55 (vo sin55/9.8)
50 (9.8) = vo^2( cos55)(sin55)
vo = sqrt[ 50(9.8) /( cos55)(sin55)]
v0 =32.29 m/s
I would start by writing an equation that relates the projectiles horizontal coordinate ("x") to its vertical coordinate ("y").
We know that
x = (Vx_initial)(t)
and:
y = (Vy_initial)(t) - ½gt²
From the first equation we get "t = x/Vx_initial". Substitute that into the 2nd equation:
y = (Vy_initial(x/Vx_initial) - ½g(x/Vx_initial)²
= (x)tanθ - ½x²/Vx_initial²
Let "V" be the initial speed (the quantity you want). Then Vx_initial = Vcosθ. Substitute:
y = (x)tanθ - ½x²/(V²cos²θ)
Now, we want the projectile to go forward 50 meters, and up 30 meters. So substitute "50 meters" for "x" and "30 meters" for "y" in the above equation, and then just solve for "V".