A body is projected vertically upward with speed 40 m / s. The distance travelled by body in the last second of upward joumey is [take g=9.8 m / s 2 and neglect effect of air resistance]

Question

A body is projected vertically upward with speed 40 m / s. The distance travelled by body in the last second of upward joumey is [take g=9.8 m / s 2 and neglect effect of air resistance] (A) 4.9 m (B) 9.8 m (C) 12.4 m (D) 19.6 m

Tardigrade Admin · Accepted Answer

As the motion under gravity is symmetric, so distance travelled in last second of ascent is equal to first second of descent.

t = 1 s

(

1^{st}

second)

- x_{2} = u t - \frac{1}{2} g \times 1^{2}

x_{2} = \frac{1}{2} \times 9.8 \times 1^{2} (∵ u = 0)

\Rightarrow x_{2} = 4.9 m

This distance is constant for every body thrown with any speed.

Q. A body is projected vertically upward with speed 40m/s. The distance travelled by body in the last second of upward joumey is [take g=9.8m/s2 and neglect effect of air resistance]

4.9 m

9.8 m

12.4 m

19.6 m

As the motion under gravity is symmetric, so distance travelled in last second of ascent is equal to first second of descent. t=1s (1st second) −x2​=ut−21​g×12 x2​=21​×9.8×12(∵u=0) ⇒x2​=4.9m This distance is constant for every body thrown with any speed.

Q. A body is projected vertically upward with speed $40 m / s$ . The distance travelled by body in the last second of upward joumey is [take $g = 9.8 m / s^{2}$ and neglect effect of air resistance]