Aprojectile is thrown upward so that its distance is h=14t square
+ 476t. what is the maximum height?

y (max)  = 4046 m

Step-by-step explanation:

 h = 14*t² + 476*t   is not the equation for a projectile upward movemente, since in that case projectile will continue going up to nfinite as time goes on. We must correct the equation according to projectile shot. Then

y = h  =  - 14*t²  + 476 * t                    (1)

So we will have h maximum when V(y) = 0

then  D(y)/dt  =  - 28 *t  +  476

D(y)/dt  = 0      ⇒   - 28 *t  +  476  =  0     ⇒  t  =  476 / 28

t  = 17 sec

Then  y maximum is for  t = 17 plugging this value in equation (1) we get

y  =  - 14*t² + 476* t

y (max)  =  - 14* (17)²  +  476 * ( 17)   ⇒  y (max)  = - 4046  +  8092

y (max)  = 4046 m

answer: the correct one is a

step-by-step explanation:

true

step-by-step explanation: