Theory question 108
DynamicsAn object is projected vertically upwards, its height, h m, at time t seconds is given by h = 20t − (3/2)t2 − (2/3)t3. Find (a) the time at which it is momentarily at rest; (b) correct to two decimal places, the maximum height reached by the object
Show the model answer
Model answer
(a) v = dh/dt = 20 − 3t − 2t² = 0
2t² + 3t − 20 = 0, (2t − 5)(t + 4) = 0
t = 2.5 s (t = −4 rejected)
(b) h = 20(2.5) − 1.5(2.5)² − (2/3)(2.5)³ = 50 − 9.375 − 10.417 ≈ 30.21 m
Explanation
Velocity is dh/dt; the object is at rest (at the top) when the velocity is zero.