Question

A particle is projected such that the vertical component of its initial velocity is 30 m/s. The height traversed by the particle in the last second of its ascent is (in m). (Take g = 10 m/s²)

Answer 1

5

Answer 2

The time of ascent is calculated using $v_y = v_{0y} - gt$, setting $v_y=0$, which gives $t_{ascent} = 3$ s. The height traversed in the last second (the 3rd second) can be found by calculating the difference in height at $t=3$ s and $t=2$ s. Using $y(t) = v_{0y}t - \frac{1}{2}gt^2$, we get $y(3) = 30(3) - \frac{1}{2}(10)(3)^2 = 45$ m and $y(2) = 30(2) - \frac{1}{2}(10)(2)^2 = 40$ m. The difference is $45 - 40 = 5$ m.