Question

I've tried out this question so many different ways and I haven't been able to get it. I put 320 ft/s into the original equation to get t (time) and plugged that value into the derivative of the ogival which is the velocity. Help would be greatly appreciated. This is college calculus.

Answer 1

The equation is s=144t - 16t², s being the distance and V=144 ft/s

a)The maximum (or minimum) is reached when the derivative of s = 0

 s' = 144 - 32t = 0  and t= 9/2  or 4.5 second (time needed to reach the max
Now plug t= 4.5 in the original equation and you will find s = 324 ft

(Why it's a max and not a minimum: because 2nd derivative is negative and equals s" = -32)

b) Velocity of the ball: distance (up) s=320 ft, time that means:
 320 =144t - 16t² (solve for x) -16t²+144t-320 = 0, t₁ =4 and t₂ =5 (t₂ is not valid since max time to reach summit is 4.5) so t =4s , distance = 320. so the velocity at 320ft is 320/4 = 80 ft/s

c) Due to the symmetry of the parabola, the velocity at 320 ft on its way down = 80 ft/s