1.

Derive expression for the maximum height and range of projectile motion ​

Answer»

Answer:

Let a projectile move with a VELOCITY u which is INCLINED with the HORIZONTAL at angle of θ. The velocity after time 't' will be

v=v

x

i

^

+v

y

j

^

v=ucosθ

i

^

+(usinθ−gt)

j

^

(Since:v

y

=usinθ−gt)

At maximum height the projectile will only have horizontal COMPONENT that is

v

x

=ucosθ

v

y

2

−u

y

2

=2ay

v

y

=0(atmaxheightH)

u

y

=usinθ

ay=−gH

Puttingthesevalues,

0=(usinθ)

2

−2gH

H=

2g

u

2

sin

2

θ

Explanation:

hope it will help you follow me please


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