NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=-4.9t² +319t+290. Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? Round the answer to the nearest tenth. The rocket splashes down after seconds. How high above sea-level does the rocket get at its peak? Round the answer to the nearest tenth. The rocket peaks at meters above sea-level.
The Correct Answer and Explanation is:
We are given the height function of a rocket: h(t)=−4.9t2+319t+290h(t) = -4.9t^2 + 319t + 290h(t)=−4.9t2+319t+290
This is a quadratic equation in the form: h(t)=at2+bt+ch(t) = at^2 + bt + ch(t)=at2+bt+c
where:
- a=−4.9a = -4.9a=−4.9
- b=319b = 319b=319
- c=290c = 290c=290
1. Time of Splashdown (when the rocket hits the ocean)
The rocket splashes down when its height h(t)=0h(t) = 0h(t)=0 (sea level).
We solve: −4.9t2+319t+290=0-4.9t^2 + 319t + 290 = 0−4.9t2+319t+290=0
Using the quadratic formula: t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}t=2a−b±b2−4ac
Substitute values: t=−319±3192−4(−4.9)(290)2(−4.9)t = \frac{-319 \pm \sqrt{319^2 – 4(-4.9)(290)}}{2(-4.9)}t=2(−4.9)−319±3192−4(−4.9)(290) t=−319±101761−(−5684)−9.8t = \frac{-319 \pm \sqrt{101761 – (-5684)}}{-9.8}t=−9.8−319±101761−(−5684) t=−319±107445−9.8t = \frac{-319 \pm \sqrt{107445}}{-9.8}t=−9.8−319±107445 t=−319±327.8−9.8t = \frac{-319 \pm 327.8}{-9.8}t=−9.8−319±327.8
Now calculate both solutions:
- t1=−319+327.8−9.8=8.8−9.8≈−0.9t_1 = \frac{-319 + 327.8}{-9.8} = \frac{8.8}{-9.8} \approx -0.9t1=−9.8−319+327.8=−9.88.8≈−0.9 (Not valid, negative time)
- t2=−319−327.8−9.8=−646.8−9.8≈66.1t_2 = \frac{-319 – 327.8}{-9.8} = \frac{-646.8}{-9.8} \approx 66.1t2=−9.8−319−327.8=−9.8−646.8≈66.1
✅ The rocket splashes down after approximately 66.1 seconds.
2. Maximum Height (the peak)
The time of the vertex (maximum point for a downward-opening parabola) is given by: t=−b2a=−3192(−4.9)=−319−9.8≈32.6 secondst = \frac{-b}{2a} = \frac{-319}{2(-4.9)} = \frac{-319}{-9.8} \approx 32.6 \text{ seconds}t=2a−b=2(−4.9)−319=−9.8−319≈32.6 seconds
Now plug t=32.6t = 32.6t=32.6 into the height function: h(32.6)=−4.9(32.6)2+319(32.6)+290h(32.6) = -4.9(32.6)^2 + 319(32.6) + 290h(32.6)=−4.9(32.6)2+319(32.6)+290 =−4.9(1062.8)+10389.4+290= -4.9(1062.8) + 10389.4 + 290=−4.9(1062.8)+10389.4+290 =−5207.7+10389.4+290=5471.7 meters= -5207.7 + 10389.4 + 290 = 5471.7 \text{ meters}=−5207.7+10389.4+290=5471.7 meters
✅ The rocket peaks at approximately 5471.7 meters above sea-level.
✅ Final Answers:
- Splashdown time: 66.1 seconds
- Maximum height: 5471.7 meters
📘 Explanation:
This problem involves analyzing the motion of a rocket using a quadratic function. The height function is a downward-opening parabola (because the coefficient of t2t^2t2 is negative), which models the rise and fall of the rocket. The rocket first ascends to a maximum height (the vertex of the parabola) and then descends until it hits the ocean (when height is zero).
To find when it hits the ocean, we set the height equation equal to zero and solve for time using the quadratic formula. The positive root gives us the splashdown time.
To find the highest point, we compute the vertex of the parabola using t=−b2at = -\frac{b}{2a}t=−2ab and then substitute that time back into the height equation to determine the peak altitude.
This combines algebra and physics to describe projectile motion.