Question

Show that the maximum height and range of a projectile are u²sin²ø/2g and 4²sin2ø/g​

Answer 1

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

Explanation:

Please please please mark my answer at brain list.