Question

A particle is projected horizontally in air at a height of 25 m from the ground with a speed of 10 m/s. The speed of the particle after 2 seconds will be

Answer 1

answer : 10√5 m/s

A particle is projected horizontally in air at a height of 25m from the ground with a speed 10m/s.

for horizontal component, use formula,

$v_x=u_x+a_xt$ but we know in horizontal direction, acceleration = 0

so, $a_x=0,u_x=10m/s$

and $v_x=10m/s$....(1)

for vertical component, use formula, $v_y=u_y+a_yt$

we know, acceleration along vertical direction is equal to acceleration due to gravity .

so, $a_y=-10m/s^2,u_y=0$ and t = 2sec

so, $v_y=0-10\times2=-20m/s$....(2)

now in vector from , velocity of particle after 2sec will be, $v=10\hat{i}-20\hat{j}$

magnitude of velocity of particle , v = √{10² + (-20)²} = √(500) = 10√5 m/s