Answer:
y1 = -130 m
y2 = 130.99 m
Explanation:
For this exercise we can use energy conservation
Starting point Higher
[tex]Em_{o}[/tex] = U = mg h
.h = 215 m
Final point. In the lowest part of the trajectory
Em_{f} = Ke + U
Em_{f} = ½ k (y-45) 2 + m g y
Energy is conserved so
Em_{o} = Em_{f}
m g h = ½ k (y- 45)² + m g y
150 9.8 214 = ½ 33 (y-45)² +150 9.8 y
314580 = 16.5 (y² - 2 45 y + 45²) + 1470 y
314580-16.5 2025 = 16.5 y² - 1485 y + 1470y
281169.5 = 16.5 y² - 15 y
y² - 0.9090 y - 17040.57 = 0
We solve the second degree equation
y = [0.9090 ±√ (0.9090² + 4 17040.57)] / 2
y = [0.9090 ± 261.08] / 2
y1 = -130 m
y2 = 130.99 m
Therefore the body drops to 130.99 m