- #1
salmayoussef
- 31
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Hey all! I've tried solving this problem every way I could think of and I wanted to see if I'm doing this even remotely right. Mind helping me out?
Problem: A mountain climber is stranded on a ledge 30 m above the ground. Rescuers on the ground want to shoot a projectile to him with a rope attached it. If the projectile is directed upward at an initial angle of 55° from a horiontal distance of 50 m, derermine the intial speed the projectile must have in order to land on the ledge.
Given:
dx = 50 m
dy = 30 m
θ = 55°
g = -9.8 m/s2
Required:
Δt
V0
Not sure if I used the proper equation but: d = V * t - (1/2)(-9.8)(t)2
I tried finding Δt first by using t = dx/Vx = 50/(cos55 * V0)
After finding the time, I used it an inputed it into the equation and canceled out the V0 in the numerator and denominator then I was left with one V which I had to find by rearranging the equation.
30 m = (sin55 * V0)(50/(cos55 * V0) - 4.9(50/cos55 * V0)2
After rearranging it, I ended up with 30.06 m/s as the initial velocity. Am I using the right equation?
Homework Statement
Problem: A mountain climber is stranded on a ledge 30 m above the ground. Rescuers on the ground want to shoot a projectile to him with a rope attached it. If the projectile is directed upward at an initial angle of 55° from a horiontal distance of 50 m, derermine the intial speed the projectile must have in order to land on the ledge.
Given:
dx = 50 m
dy = 30 m
θ = 55°
g = -9.8 m/s2
Required:
Δt
V0
Homework Equations
Not sure if I used the proper equation but: d = V * t - (1/2)(-9.8)(t)2
The Attempt at a Solution
I tried finding Δt first by using t = dx/Vx = 50/(cos55 * V0)
After finding the time, I used it an inputed it into the equation and canceled out the V0 in the numerator and denominator then I was left with one V which I had to find by rearranging the equation.
30 m = (sin55 * V0)(50/(cos55 * V0) - 4.9(50/cos55 * V0)2
After rearranging it, I ended up with 30.06 m/s as the initial velocity. Am I using the right equation?