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

ruth is 6 years old

step-by-step explanation:

6 + 24 = 30,   so 6/30 = .2  

if you do 1/5 it also equals .2, so that is the answer.

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )