Question

A particle is projected horizontally from top of the building of height 20 metres with a speed 30metres per second distance of the point where particle lands on ground from its point of projection

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Answer 1

Answer:

ANSWER

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

Explanation:

hope it helps you dear

Answer 2

Given : A particle is projected horizontally from top of the building of height 20 metres with a speed 30metres per second

To Find : distance of the point where particle lands on ground from its point of projection

Solution:

Height = 20 m

Horizontal Speed = 30 m/s

Vertical Speed = 0 m/s

Vertical acceleration = g  =   m/s²

Horizontal acceleration = 0 m/s²

S = ut + (1/2)at²

20 = 0 + (1/2)10t²

=> t² = 4

=> t = 2 sec

Horizontal Distance = 20 x 2  =  40 m

Vertical Distance = 20 m

Distance = √40² + 20² = 20√5  m

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