A football player releases a ball at 35° with initial velocity of 80 ft/sec. Determine the radius of curvature of the trajectory at times t = 1 sec and t = 2 sec, where t = 0 is the time of release from the quarterback's hand. For each case, compute the time rate of change of the speed.
Book answers: at t = 1: p = 142.2 ft, at = -6.58 ft/sec^2
at t = 2: p = 149.7 ft, at = 8.75 ft/sec^2
an = v^2/ρ
at = dv/dt = acceleration in tangential direction
v = ρ(dΘ/dt)
a = dv/dt
The Attempt at a Solution
I'm really confused about how to tackle a problem like this that involves specific times. In class and in all the homework up to this point we haven't had an example of how to approach these types of problems, but I'm guessing there's a derivative or integral that I'm not seeing or understanding that's killing me, but the closest thing I could come up with is:
magnitude of acceleration = g = 32.2 ft/sec^2
v = ρ(dΘ/dt), which I split into: v dt = ρ dΘ.
I then made an integral: ∫(from 0 to t) 80 dt = ρ ∫(from 0 to 35*π/180) dΘ.
This gave me 80t = (.6109)ρ, when t = 1 gives me ρ = 131 ft.
I then tried solving for an with v^2/ρ, which gave me an = 48.87, which can't be possible as it makes magnitude a = g false...
Thanks in advance.