Question

A projectile is fired to have maximum horizontal range r. The maximum height reached by the projectile in this case is

Answer 1

Answer:

Given:

A projectile is fired to have maximum range "r".

To find:

Maximum height

Concept:

We know that the formula for the range of a projectile =

u²sin(2θ)/g

So for max range , sin(2θ) should have max value = 1.

Hence 2θ = 90°

=> θ = 45°.

so angle of Projection is 45°

Formulas used:

Range = u²sin(2θ)/g

Max height = u²sin²(θ)/2g

Calculation:

Max Range = u²sin(90°)/g

=> r = u²/g ...........(i)

Again , max height = u²sin²(θ)/2g

=> H (max.) = u²sin²(45°)/2g

=> H (max.) = u²/2g × (1/√2)²

=> H (max.) = u²/2g × ½

=> H (max.) = u²/4g

=> H (max.) = ¼ (u²/g)

=> H (max.) = r/4 .

Answer 2

$\Large{\underline{\underline{\red{\sf{Answer :}}}}}$

$\Large \displaystyle {\underline{\boxed{\sf{h_{max} \: = \: \frac{R}{4}}}}}$

$\large{\underline{\underline{\red{\sf{Step-By-Step-Explanation :}}}}}$

The formula for Horizontal Range is :-

$\Large \displaystyle {\boxed{\sf{\star \: R \: = \: \frac{u^2 Sin2 \theta}{g}}}}$

For Maximum Range value of theta is 45°

$\rightarrow \displaystyle {\sf{R \: = \: \frac{u^2 Sin 2(45)}{g}}}$

$\rightarrow \displaystyle {\sf{R \: = \: \frac{u^2 Sin (90)}{g}}}$

$\rightarrow \displaystyle {\sf{R \: = \: \frac{u^2 (1)}{g}}}$

$\Large \implies \displaystyle {\boxed{\sf{R \: = \: \frac{u^2}{g}}}}$

And Formula for maximum height is :-

$\Large \displaystyle {\boxed {\sf{\star \: h \: = \: \frac{u^2 Sin^2 \theta}{2g}}}}$

For Maximum height theta is equal to 45°

So,

$\rightarrow \displaystyle {\sf{h \: = \: \frac{u^2 Sin (45^{\circ})}{2g}}}$

$\rightarrow \displaystyle {\sf{h \: = \: \frac{u^2 (\frac{1}{\sqrt{2}})^2}{2g}}}$

$\rightarrow \displaystyle {\sf{h \: = \: \frac{u^2 }{4g}}}$

$\Large \implies \displaystyle{\boxed{\sf{h \: = \: \frac{R}{4}}}}$