Question

A particle is thrown with a any velocity vertically upward , the distance travelled by the particle in first second of its decent is

Answer 1

When a particle is thrown vertically upwards, its velocity goes on decreasing until it becomes zero at Maximum Height.


At Maximum Height, velocity of the ball is zero. Gravitational Acceleration g is still acting downwards.


The particle now accelerates downward. We can easily find the distance covered in the first second of its descent.


Our data would be:

$\text{Initial Velocity} = u = 0 \, \, m/s \\ \\ \text{Acceleration} = g = 9.8 \, \, m/s^2 \\ \\ \text{Time} = t = 1 \, \,s$


We can apply equations of motion. We will use:



$\displaystyle s = ut + \frac{1}{2}at^2 \\ \\ \\ \implies s = (0)(1) + \frac{1}{2} \times 9.8 \times 1^2 \\ \\ \\ \implies s = 0 + 0 + 4.9 \, \, m \\ \\ \\ \implies \boxed{\bold{s=4.9 \, \, m}}$



Thus, The distance travelled by the particle in the first second of its descent is 4.9 metres

Answer 2

Answer:

g/2 (option 2)

Explanation:

this will help u...as I think..