206k views
2 votes
Helppp !!! Giving 20 pts !!

Helppp !!! Giving 20 pts !!-example-1

2 Answers

5 votes

Answer: 6.12 seconds

Explanation:

To find when the object strikes the ground, we need to find the time t when the height h(t) is equal to 0 (since the ground is at height 0). We have the equation:

h(t) = -4.9t² + 9.31t + 239.12

To find the time when the object strikes the ground, we need to find the value of t when h(t) = 0:

0 = -4.9t² + 9.31t + 239.12

This is a quadratic equation of the form at² + bt + c = 0, where a = -4.9, b = 9.31, and c = 239.12. We can solve this equation for t using the quadratic formula:

t = (-b ± √(b² - 4ac)) / 2a

Plugging the values of a, b, and c into the formula, we get:

t = (-(9.31) ± √((9.31)² - 4(-4.9)(239.12))) / (2(-4.9))

t = (-9.31 ± √(86.6161 + 4694.336)) / (-9.8)

t = (-9.31 ± √(4780.9521)) / (-9.8)

t ≈ (-9.31 ± 69.14) / (-9.8)

We will have two possible values for t:

t₁ ≈ (-9.31 + 69.14) / (-9.8) ≈ 6.12 (rounded to two decimal places)

t₂ ≈ (-9.31 - 69.14) / (-9.8) ≈ 8.00 (rounded to two decimal places)

Since the height function describes the motion of the ball from a platform, we can discard the negative solution as it represents an invalid time before the ball is launched. Thus, the object strikes the ground at approximately t ≈ 6.12 seconds.

User Rico Harisin
by
8.6k points
1 vote

To determine when the ball strikes the ground, we need to find the value of t when the height h(t) is 0.

The equation for the ball's height h at time t seconds after launch is:

h(t) = -4.9t^2 + 9.31t + 239.12

Setting h(t) to 0, we get:

0 = -4.9t^2 + 9.31t + 239.12

We can solve this quadratic equation using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = -4.9, b = 9.31, and c = 239.12.

User Louis W
by
8.5k points

No related questions found