Question

Derive
an expression for
(a) Maximum Height (h) ​

Answer 1

Answer:

Explanation:

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

​

​