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.

Ive tried out this question so many different ways and I havent been able to get it I put 320 fts into the original equation to get t time and plugged that valu class=

Respuesta :

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