Question

A particle is projected horizontally at 15m/s from a height of 20m. calculate the horizontal distance covered by the particle just before hitting the ground.

Answer 1

${\huge{\bold{\blue{\underline{♡AnSwEr-♡}}}}}$

Let R represent the horizontal distance covered at time t.

Using h = ut + 1/2 gt^2

but Uv = 0

h = 1/2gt^2

20 = 1/2× 10×t^2

20 = 5t^2

t^2 = 20/5 = 4

t = 2s

Range(R) = Ux × t

= 15 x 2

= 30m

or

_____

R = u x √2h/g

= 15× √2×20

-------------

10

= 15 x √4

= 15 x 2

= 30m

Answer 2

Let R represent the horizontal distance covered at time t.

Using h = ut + 1/2gt2

but Uv = 0

h = 1/2gt2

20 = 1/2×10×t2

20 = 5*t2

t2 = 20/5 = 4

t = 2s

Range(R) = Ux x t

= 15 x 2

= 30m

or

R = u x 2hg−−√

= 15 x 2×2010−−−−√

= 15 x 4–√

= 15 x 2

= 30m