Question

of the horizontal, maximum projection is of a projectele s R & the is R height is h, show that the velocity of 2g (H + R^2÷16 H)^1/2
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Answer 1

Answer:

Correct option is

B

2gh+

8H

R

2

g

Let the velocity of the projection is 'V' and angle of projection "θ" with horizontal

At maximum height H,v

y

=0

Using Newton's third equation of motion

v

2

−u

2

=2as

u=0,u=vsinθ;a=−g;s=H

0−v

2

sin

2

θ=2×(−g)H

vsinθ=

2gH

⟶1

Also the half time of flight-from A to B and from B to C

Using Newton's second equation of motion from B to C

u=o;t=

2

T

;S=−H,a=−g (Here T is total time of flight)

S=ut+

2

1

at

2

−H=0+

2

1

(−g)(

2

T

)

2

T=

g

8H

Range (R)=vcosθ×T

R=vcosθ×

g

8H

⟶2

vcosθ=R

8H

g

⟶3

Squaring and adding 1 and 2

v

2

=2gH+

8H

R

2

g

v=

2gH+

8H

R

2

g