Question

show that the maximom height
by a projectile is
reached
Imax=(rosin)
ago​

Answer 1

Answer:

Step-by-step explanation:

In projectiles Range, ( R )= `(u^2sin2theta)/g`

Maximum height (H)= `(u^2sin^2theta)/(2g)`

When `theta=45^@`

`impliesR=R_(max)=(u^2sin90^@)/g`

`R_(max)=u^2/g`

Maximum height

`H_(max)=(u^2sin^2(45^@))/(2g)`

`=(u^2(1//sqrt2)^2)/(2g)=u^2/(4g)`

`H_(max)=1/4R_(max)`

`therefore` When `theta=45^@` maximum height reached is one quarter of maximum range.