- #1
NotFrankieMuniz
- 3
- 0
Homework Statement
:[/B]A projectile is given initial velocity 80 m/s ( V0 ) at angle 60° above the horizontal. Find the time it takes to reach to the highest point and find the maximum height. (g = -10 m/s2)
Homework Equations
:[/B]1. [itex]y_{final} = y_{initial} + v_{y_{initial}}t + \frac {1}{2}gt^2[/itex]
2. [itex]v_{final}^2 = v_{y_{intial}^2} + 2g(y_{final} - y_{intial})[/itex]
The Attempt at a Solution
:[/B]
The first thing I did was to find the velocity of the y-component ( [itex]v_{y_{initial}})[/itex]:
[itex]v_{y_{initial}} = (80 m/s)({\sin 60}^{\circ})[/itex]
[itex]v_{y_{initial}} = 69 m/s[/itex]
Since I know that [itex]v_{final}^2[/itex] is 0 m/s at the highest point, I can use Equation 2 to figure out the max height. With [itex]v_{y_{initial}^2}[/itex] known and [itex]t[/itex] unknown:
[itex]v_{final}^2 = v_{y_{intial}^2} + 2g(y_{final} - y_{intial})[/itex]
[itex]0 = v_{y_{intial}^2} + 2gy[/itex]
[itex]y = \frac{-v_{y_{intial}^2}}{2g}[/itex]
[itex]y = \frac{-(69 m/s)^2}{2(-10 m/s^2)}[/itex]
[itex]y = 238 m[/itex]
With the max height known, I can use Equation 1 to get the time to reach the max height:
[itex]y_{final} = y_{initial} + v_{y_{initial}}t + \frac {1}{2}gt^2[/itex]
[itex]238 m = (0m) + (69 m/s)t + \frac {1}{2}(-10m/s^2)t^2[/itex]
[itex](5m/s^2)t^2 - (69 m/s)t + (238 m) = 0[/itex]
From there, I used the quadratic formula where a = 5 m/s2, b = -69 m/s, and c = 238 m
The results that I got were t = 175 s and t = 170 s, which doesn't seem right.
Anyone care to point me to the right direction?
Last edited: