A diver on a platform 40 feet in height jumps upward with an initial velocity of 5ft/sec. His height in h feet after t seconds is given by the function h = -16t^2 + 5t + 50.

Question

A diver on a platform 40 feet in height jumps upward with an initial velocity of 5ft/sec. His height in h feet after t seconds is given by the function h = -16t^2 + 5t + 50. A. What is the maximum height? B. How long will it take him to reach the surface of the water?

Answer 1

A. Maximum height at t = -b/2a, or t = -5/-32

B. 0 = -16t² + 5t + 50, can you solve for t and answer?

Answer 2

Odd question. The platform is obviously 50 ft. tall at t=0.

Regardless, graph the equation or use -b/2a to find t at the vertex:

-5/-32=.15625. Plug this in the original quadratic to find that at .15625 seconds, the diver is about 50.391 feet above the water.

Use the quadratic formula or graph the function to find the values where (h(t))=0. The only positive solution is where t=1.931.

The quadratic formula, if you don't know how to use it, is something you **need to memorize** because quadratic equations are so prevalent in our world. Here's a video explaining it: https://www.youtube.com/watch?v=i7idZfS8t8w

I hope this helps, and have fun mathing!

Answer 3

-16t^2 + 5t +50

take the derivative and set equal to zero, solve for t, that's the time of max height.

-32t +5 = 0

t = 5/32 secons

plug into the original equation to solve for max height

-16(5/32)^2 + 5(5/32) + 50 =

-25/64 +50/64 +50 =

25/64 +50 = about 50.39 feet. He only goes up about 4/10 of a foot = 4/10x12 = less than 5 inches in

about 1/6 of a second

he hits water when h=0 = -16t^2 +5t +50. multiply by -1

16t^2 -5t -50 = 0 Use the quadratic formula t=-b/2a + (1/2a)sqr(b^2-4ac) (ignore neg. sqr)

t = 10/32 + (1/32)sqr(25+64(50))= about (5+56.8)/32 = 1.93 seconds

There's an ambiguity in the problem though. It says the platform is 40 feet high.

Yet the equation has a constant term of 50. The constant term is supposed to be the initial height.

Is it 40 or 50? IF 40, then the time is less to reach the water. time to reach max height is still the same,

either way. Maybe the platform is 40 feet high, but the water is somehow another 10 feet below that?

Or maybe there was a typo? As 4 and 5 are close on the keyboard?

the general equation is h(t) = (a/2)t^2 + vot + ho where a= deceleration due to gravity = -32 feet per second per second, vo = initial velocity and ho = initial height.