Question

Suppose h(t)=-5t^2+10t+3 is the height of a diver above the water(in meters), t seconds after the diver leaves the springboard. When does the diver hit the water? and at what time on the divers descent toward the water is the diver again at the same height as the springboard?

Answer 1

Answer:

See Below:

Step-by-step explanation:

Hit Water:

He hits the water when his height, h(t), is 0 meters above water.

0 = $-5t^2+10t+3$

This is no factorable so we use the quadratic formula.

(-b±√(b²-4ac))/2a

(-10±√(10²-4*-5*3))/2*-5

(-10±√(100+60))/-10

(-10±√160)/-10

(-10±√16*10)/-10

(-10±4√10)/-10

(-10±4√10)/-10         Seperate to two equations

(-10+ 4√10)/-10                          and                (-10- 4√10)/-10

(-10+12.65)/-10                                                 (-10-12.65)/-10  

-.265 Not Solution                                          2.265 Seconds

Becasue there is no negative time

The diver hits the water at 2.265 seconds

For when he is at the same height as the spring board, we need to find the height of the spring board, and we do that by substituting 0 for t because we need the height 0 seconds after he jumps.

h(0) = -5*0^2 +10*0 +3

h(0) = 3   The height of the spring board is 3 so we need to know when he is at 3 meters again.

3= -5t^2 + 10t + 3       Subtract 3 from both sides

0= -5t^2 +10t

0= -5(t^2-2t)       Factor out -5 and divide both sides by -5

0= t^2-2t    

0= t(t-2)      Factor out t.

Solve.

t= 0               t-2=0

                     t= 2

He will be at the same height as the spring board at t= 2 seconds