Question

A body is projected from a cliff at an angle of 60°
with horizontal with velocity 40 m/s. It reaches
ground in 5root3 s, height of the cliff is​

Answer 1

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

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Answer 2

Answer:

angle=60°

velocity =40 m/s.

time =5√3s

height=?

ut+1/2gt^2

=40*5√3+1/2*9.8*5√3^2

=419.9