Question

A particle is projected horizontally from the top of a building of height 20 m with a speed of 30 m/s. Distance of point where particle lands on ground from its point of projection is​

Answer 1

Explanation:

R = 40 m ; it was projected horizontally

So that horizontal velocity =

t

R

t = time ti free fall a height 20 m

↓

as initial vertical velocity = 0

Given by h=

2

1

9t

2

we have t =

9

2h

t=

10

2×20

= 2sec

& vertical velocity gained in these 2 sec

v

y

=u+9t=0+10×2=20 m/s

Horizontal velocity v

x

=

t

R

=

2

40

=20m/s (constant)

⇒ speed =

v

y

2

+v

x

2

=20

2

m/s