Examkin

Soru

Zorluk: OrtaQuadratic Equations and the Quadratic Formula

A projectile is launched vertically upward from an initial height of 55 meters. Its height, h(t)h(t) in meters, tt seconds after launch is given by the function h(t)=−4.9t2+19.6t+5h(t) = -4.9t^2 + 19.6t + 5. To the nearest tenth of a second, how many seconds after launch does the projectile reach a height of 1515 meters on its way down?

Cevap: 3.4 seconds

Cevap

To the nearest tenth of a second, the projectile reaches a height of 1515 meters on its way down at 3.43.4 seconds.
The correct answer is 3.43.4 seconds. Setting the height equation h(t)=15h(t) = 15 yields −4.9t2+19.6t−10=0-4.9t^2 + 19.6t - 10 = 0. Solving this quadratic equation via the quadratic formula gives two solutions: t≈0.6t \approx 0.6 seconds and t≈3.4t \approx 3.4 seconds. The projectile travels upward first, passing the 1515-meter mark at 0.60.6 seconds, and then descends, passing the 1515-meter mark again at 3.43.4 seconds.

Adım Adım Çözüm

1
Set up the quadratic equation by setting the height function h(t)h(t) equal to 1515.
−4.9t2+19.6t+5=15-4.9t^2 + 19.6t + 5 = 15
This allows us to find the specific values of time tt when the height of the projectile is exactly 1515 meters.
2
Rearrange the quadratic equation into the standard form at2+bt+c=0at^2 + bt + c = 0 by subtracting 1515 from both sides.
−4.9t2+19.6t−10=0-4.9t^2 + 19.6t - 10 = 0
Writing the equation in standard form is necessary before applying the quadratic formula.
3
Substitute the coefficients a=−4.9a = -4.9, b=19.6b = 19.6, and c=−10c = -10 into the quadratic formula.
t=−19.6±(19.6)2−4(−4.9)(−10)2(−4.9)t = \frac{-19.6 \pm \sqrt{(19.6)^2 - 4(-4.9)(-10)}}{2(-4.9)}
Since the quadratic equation has non-integer decimal coefficients, using the quadratic formula is the most reliable method to solve for the roots.
4
Simplify the discriminant and calculate the two values of tt.
t≈0.6t \approx 0.6 and t≈3.4t \approx 3.4
The discriminant is 19.62−196=188.1619.6^2 - 196 = 188.16. Taking the square root gives 188.16≈13.72\sqrt{188.16} \approx 13.72, resulting in two real roots.
5
Determine which root corresponds to the projectile's motion on the way down.
t≈3.4t \approx 3.4 seconds
The smaller root (0.60.6 seconds) represents the first time the projectile reaches 1515 meters while ascending. The larger root (3.43.4 seconds) represents the time the projectile passes 1515 meters while descending.

Anahtar Kavram

Solving quadratic equations with decimal coefficients using the quadratic formula and interpreting the physical context of the roots.
Bu soruyu puanla

Konular

Quadratic Equations and the Quadratic Formula

Bu konudan daha fazla soru