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# Math Quadratic Formulas?

A rock is thrown directly upward with an initial velocity of 136 feet per second from a cliff 295 feet above a beach. The height of the rock above the beach h after t seconds, is given by the equation h=−16t2+136t+295.

When will the rock hit the beach? If you have more than one answer, separate them with a comma. If necessary, round to 2 decimal places.

### 3 Answers

- Wayne DeguManLv 72 months ago
Relative to the beach we have a = -g => -32

Velocity is 136 and displacement at t = 0 is 295

Using s = ut + (1/2)at² we get:

h = 136t + (1/2)(-32)t² + 295

i.e. h = -16t² + 136t + 295...confirming what you state.

The beach is at ground level, so the rock hits the beach when h = 0

i.e. when -16t² + 136t + 295 = 0

or, 16t² - 136t - 295 = 0

Using the quadratic formula we have:

t = [136 ± √(-136)² - 4(16)(-295)]/32

so, t = (136 ± 16√146)/32

Hence, t = -1.79 or t = 10.29

Therefore, the rock hits the beach at t = 10.29 seconds

:)>

- husoskiLv 72 months ago
If the rock is thrown directly upward by a person standing on a cliff, it will land where the person is standing--on the cliff. It never hits the beach.

[Edit: Some extra unreality of this "real-world problem" iss that equation means that the rock was thrown at 136 ft/s. That's over 92 MPH, a major league fastball, not an underhand toss with arm extended over the edge! I wonder if any MLB pitchers can throw that fast straight up...]

If you're willing to ignore that, and assume that "directly upward" means "almost directly upward so that it misses the cliff on the way down but still stays over the beach and doesn't land in the water", then the the height h will be zero when the rock hits the beach. Solve for t:

h = 0

-16t² + 136t + 295 = 0 . . . substitute given expression for h

That's a quadratic equation in standard ax² + bx + c = 0 form, with t for the independent variable (instead of x), a=-16, b=136 and c=295. The quadratic formula is:

t = [-b ± √(b² - 4ac)] / (2a)

The rest is mostly just punching calculator buttons, except...

They already tried to mislead you with "directly upward". Then they try again with the "If you have more than one answer..." bit. The rock can't hit more than once and still be following that equation for height vs. time. It wasn't in flight at all before being thrown, so there are no valid negative answers for t. It won't continue following the equation after it hits, so there can't be two positive solutions either. So the only valid answer to the problem is the smallest positive solution to that quadratic equation.

Hint: However, if you get two positive solutions in this particular problem, you've done something wrong!