Question

A body is projected horizontally height of 5m .It reaches the ground at horizontal distance of 10m . The speed of the particle when it reaches the ground ???​

Answer 1

it is base on horizontal projectile motion. we know, in case of projectile motion, horizontal component of acceleration is zero and speed along horizontal direction always remains same.

if we assume body is projected horizontally with velocity u .

it means, Horizontal component of velocity, u.

and vertical component of velocity = 0

now, horizontal distance covered in time t = 10m

so, 10 = ut......(1)

again, vertical distance covered in time t, = 5m

so, 5 = 0 + 1/2 gt² [ from s = ut + 1/2 at² ]

from equation (1),

5 = 1/2 g(10/u)²

10/g = (10/u)²

if g = 10m/s²

10/10 = 1 = (10/u)² => u = 10m/s

hence, initial speed of particle is 10m/s

now, time taken to reach the ground , t = u/g = 1sec

so, velocity after time t = 1sec along horizontal , vx = u = 10m/s [ horizontal component doesn't change ]

velocity after time t = 1sef along vertical , vy = -gt = -10m/s

hence, velocity of particle when it reaches the ground , v = 10i - 10j

so, speed of particle , |v| = √(10² + 10²) = 10√2 m/s

Answer 2

Answer:

10√2 m/s

that's the answer