Question 5
An object is thrown vertically upwards with a velocity of 30 m s-1. Its velocity–time graph, up to the point where it reaches its maximum height, is shown below. (g = 10 m s-2)
| t (s) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| v (m s⁻¹) | 30 | 20 | 10 | 0 |
What is the maximum height the object rises to?
The right answer is the one marked above.
45 m. The maximum height is the area of the region under the graph. The velocity takes 3 s to fall from 30 to 0 (30, 20, 10, 0):
height = ½ × 30 m s-1 × 3 s = 45 m
The average velocity route gives the same answer: (30 + 0) ÷ 2 = 15 m s-1, multiplied by 3 s gives 45 m.
90 m comes from treating the velocity as 30 m s-1 throughout — but it falls away steadily and is zero at the top.