he function h defined by h(t)=(49+4.9t)(10−t) models the height of an object t seconds after it is dropped from a helicopter.Find or approximate the time when the object hits the ground.

Question

please help!

Answer 1

Hi Claire,

The question is asking you to use the function you were given to identify how long it takes for the object to hit the ground. In other words, it is asking you to find the value for t when h(t) = 0. Here h(t) models the height above the ground.

Using this info, we can set up the following problem:

1. Let h(t) = 0:
2. h(t) = (49+4.9t)(10−t) = 0
3. Solve for values of t that result in h(t) = 0
4. Since there are two factors, there are two solutions. One solution comes from solving 49+4.9t = 0. The other comes from solving 10-t = 0. This is because any value of t that solves either of these equations will yield a 0 in h(t) = (49+4.9t)(10−t) = 0
5. Solve for 49 + 4.9t = 0
6. Factor out 4.9 to simplify: 4.9 (10 + t) = 0
7. Divide both sides by 4.9 to get: 10 + t = 0
8. t = -10
9. Solve for 10-t = 0
10. t = 10
11. Mathematical solutions here (roots) are: t = -10 and t = 10
12. Since the problem asks you to identify the time it takes for an object to hit the ground, the solution must be positive (no negative time or time travel allowed here). Thus, the answer must be t = 10.
13. Only realistic solution: t = 10 seconds

I hope this helps!

Answer 2

h= (49 +4.9t)(10-t)

= -4.9t^2 + 490 + 49t -49t

= -4.9t^2 +490

h= 0 = -4.9t^2 +490

4.9t^2 = 490

t^2 = 490/4.9 = 100

t = sqr100

t = 10 seconds to reach the ground

Answer 3

When the object hits the ground, then the height h would be zero. so make h(t) = 0 and solve for the time "t":

0 = (49 + 4.9t)(10 - t)

Here we have two things "49 + 4.9t" and "10 - t" and we are multiplying them together and getting zero as the result. If a•b = 0 then doesn't either a = 0 or b = 0? That's the only way to get zero, right?

So make each part equal to zero:

Part 1 equals zero:

49 + 4.9t = 0

4.9t = -49

t = -10

Part 2 equals zero:

10 - t = 0

10 = t

Since we will not be going backwards in time to get -10 seconds, we can throw that answer out. The result is that the object hits the ground in 10 seconds.

Answer 4

It's already factored. Set each factor = 0, solve for t.

(49+4.9t) = 0

4.9t = -49

t = -49/4.9 = -10

AND

(10−t)= 0

-t = -10

t = 10

Drop the -10 because that's time in the past.

It will hit the ground 10 seconds after it is dropped.

Answer 5

The object hits the ground when h(t)=0 Set this equal to zero and solve.

0=(49+4.9t)(10-t) Set these factors equal to 0

49+4.9t=0 or 10-t=0 Solve

49=-4.9t t=10

t=-10

Since time cannot be negative for a work problem, t=10 seconds is the answer.