I would like help with a quadratic equation story problem. It uses h(t)=v t - 16t^2. Can anybody help me?

Question

A boat in distress launches a flare straight up with a velocity of 190 feet per seconds. Ignoring the height of the boat, how many seconds will it take for the flare to hit the water? What is the maximum height reached by the flare?

I did this so far.

h(t) = v t - 16t^2
h= 190t-16t^2.

I'm not sure where to go from here.

Thanks to everyone who answered although the most detailed answer was the best. In the meantime while I waited for answers, I did figure this out finally and I got it right. Thanks. This did help.

Answer 1

The height when it and hits back the water is equal to zero.
So: 0 = -16t^2+190
Apply quadratic formula. You get t = 11.88 or t = 0.00526 which you can take as 0. t=0 is the time when it leaves the boat while t = 11.88 is the time when it hits back the water

The flare undergoes projectile motion. At its highest point, the velocity is zero. So:
h = -16t^2

The time to reach the highest point is half of the time it takes to hit the water. So its t = 5.94s
h = -16 (5.94)^2 = -564.5376. Just ignore the negative sign. The highest height is 564.5376 ft.

Answer 2

h(t) = v t - 16t^2
h = 190t -16t^2
190t2 -16t2^2 = 0
- t2(16t2 - 190) = 0
t = 0, 11.875 s
t1 = t2/2 = 5.9375 s
h1 = 5.9375(190 - 16*5.9375)
h1 = 564.0625 ft ≈ 564 ft

Answer 3

h(t) = vt -- 16t^2
when it strikes water h(t) = 0 = 190t -- 16t^2 giving t = 0 (starting), 190/16 seconds
so it strikes water after 190/16 sec = 11 7/8 sec
maximum height reached = h(95/16) = 190(95/16) --16(95/16)^2
= 1128.125 -- 564.0625
= 564.0625 ft