The Rocket's Turning Point
A model rocket is launched from a 2 m platform. Its height in metres after t seconds is modelled by h(t) = −5t² + 20t + 2. A student is asked to find the maximum height, the time it hits the ground, and the speed at impact. The student differentiates to get h′(t) = −10t + 20, sets it to zero, and finds t = 2. She then computes h(2) = 42 m as the maximum height. For the landing time she solves −5t² + 20t + 2 = 0 and gets t = (20 ± √(400 + 40))/10, giving t ≈ 4.19 s. She then substitutes t = 4.19 into h′(t) to obtain the impact speed.
1. Verify the maximum height and explain why setting h′(t) = 0 gives a maximum rather than a minimum here.
Read the worked answer
h′(t) = −10t + 20 = 0 gives t = 2. Then h(2) = −5(4) + 20(2) + 2 = −20 + 40 + 2 = 22 m, not 42 m - the student made an arithmetic slip. To confirm it is a maximum, use the second derivative h″(t) = −10 < 0, so the curve is concave down and t = 2 is a maximum. Equivalently, the parabola opens downward because the coefficient of t² is negative. So the true maximum height is 22 m at t = 2 s.
2. Find the correct landing time and evaluate the student's method for the impact speed. Is substituting the landing time into h′(t) valid?
Read the worked answer
Solve −5t² + 20t + 2 = 0, i.e. 5t² − 20t − 2 = 0. t = (20 ± √(400 + 40))/10 = (20 ± √440)/10. √440 ≈ 20.976, so t ≈ (20 + 20.976)/10 ≈ 4.10 s (the negative root is rejected). The student's 4.19 s came from an arithmetic error under the root. Her method of substituting the landing time into h′(t) is valid: h′(t) is the instantaneous vertical velocity, so h′(4.10) = −10(4.10) + 20 ≈ −21 m/s. The negative sign indicates downward motion; the speed is about 21 m/s. Substituting into the derivative is legitimate because the derivative gives velocity at any time, including impact.