Question

. If a particle is thrown up with some velocity (more
than 10 m/s) then distance travelled by the particle
in first second of its descent will be

(1) g
(2)g/4
(3)g/2
(4) Zero
​

Answer 1

$\green{SOLUTION:}$

⇒In the first second of that particle’s descent, it will start from rest, having stopped at its highest point, and travel a distance “S” under gravitational acceleration back toward Earth.

⇒S= ut + 1/2 gt^2

S is the distance the particle will fall.

u is initial velocity at the highest point, zero.

t is the time of fall, 1 second.

g is the acceleration of gravity near Earth, or 9.8 m/s^2.

S = 0 (1) + 1/2 (9.8m/s^2) * (1 s)^2

S = 0 + (4.9 m/s^2) * (1 s^2)

S = 4.9 meters