Question

The maximum height of the projectile is a) u2sin2 θ/g b) u2sin2 θ/g c) u2sin2 θ/2g d) u2sin2θ/2g. for 11th class ​

Answer 1

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