So just looked back at one of the chapter 1 problems (Example 1.14) and I understand the turbo-solved method. My question is, why am I not getting the same answer by using the range formula R = [(vo^2)xsin(2theta)]/g?
If I just put in v initial as 15 m/s - I get 22.5 as the range...
How do you figure that v
initial is 15 m/s? That might be the problem. They don't tell you the initial speed.
If you want to double check the turbo answer, then we can get v from the answer we have. In the x-direction it covers 44.1 m in 3 s, so it's average speed is 15. In the y-direction it climbs for 1.5 s, so from v
t = at + v
o we know that its initial speed is 15 m/s. It makes sense because at 45-degrees, v
y initial must equal v
x initial. Using Pathagorean math, we find that the v
total initial is the square root of (15^2 + 15^2), which is the square root of 450 which should be about 21.
Plugging that into the range equation gives,
R = [(vo^2)xsin(2theta)]/g
R = [(21^2)xsin(90)]/10
R = [450x1]/10 = 450/10 = 45
The answers match (if you ignore the rounding in my math).
