Question

Check the correctness of the eqation
H=(v^2sin^2 thetta ) / 2 g. where H= maximum height v= initial velocity theetta= angle of projection. g = acceleration due to gravity​

Answer 1

Explanation:

Show that the maximum height and range of a projectile are

2g

u

2

sin

2

θ

and

g

u

2

sin2θ

respectively

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ANSWER

ux=ucosθax=0

uy=usinθay=−g

atmax.height

Vg=0

⇒0−usinθ=gt

⇒t=

g

usinθ

∴H=ugt−

2

1

ayt

2

⇒H=

g

usinθ

.usinθ−

2

1

×g×

g

2

u

2

sin

2

θ

∴H=

2g

u

2

sin

2

θ

Whenballcomestogoundt=

g

dusinθ

∴Range=u

x

.T+

2

1

a×T

2

=ucosθ×

g

2usinθ

∴Range=

g

u

2

sin

2

θ

solution