Question

A particle is projected in vertically upward directions.find distances travelled by the particle in last second in upward motion?

Answer 1

Question #27624

A particle is projected in vertically upward directions. find distances travelled by the particle in last second in upward motion?

Solution:

An equation of equally decelerated motion is:

$$S = v_1 t - \frac{1}{2} g t^2$$

were $v_1$ is the velocity of the particle on the beginning of last second

Such as

$$v = v_1 - g t$$

the final velocity is equal to zero

$$v_1 = g t$$ $$S = g t * t - \frac{1}{2} g t^2 = \frac{1}{2} g t^2$$

The distance is:

$$S = \frac{1}{2} g t^2,$$

were $g$ is the acceleration due the gravity, $t$ is the time (1 sec)

$$S = \frac{1}{2} * 9.8 * 1 = 4.9 \, m$$

Answer: 4.9 m.