Question

1. What will be the range on maximum height?
2. What will be the height on maximum range?
3. At what angle height will be maximum?

please explain these questions.

Answer 1

What is the angle of projection for which the range and maximum height become equal?

Range of projectile (R)={ (vi^2) (Sin 2x) } / g ………(1)

height of projectile (h)={ (vi^2) (Sin x)^2 } / 2g ………(2)

So According to required condition:

R=h

{ (vi^2) (Sin 2x) } / g = { (vi^2) (Sin x)^2 } / 2g

by solving:

2 Sin 2x = (Sin x)^2 ……(3)

As Sin 2x = 2 (Sin x) (Cos x)

now (3) becomes:

2 { 2 (Sin x) (Cos x) } = (Sin x) (Sin x)

4 (Sin x) (Cos x) = (Sin x) (Sin x)

4 (Cos x) = (Sin x) ……..(4)

Dividing both side by “Cos x” to (4), we get

4= (Sin x) / (Cos x)

Since : Tan x= (Sin x) / (Cos x)

4= Tan x

Or

Tan x = 4

x= arc Tan (4)

x= 75.96°

x= 76° (Approx…)

Hence the angle of projection for which the range and maximum height become equal is 76° . (Answer)

Answer 2

Hope u like my process
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=> greatest height

$= \bf \: \frac{ {(u \sin( \theta )) }^{2} }{2g}$
=> greatest range = $\frac{ {u}^{2} \sin(2 \theta ) }{g}$

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1)

=> so maximum height will be when
$\sin( \theta ) =1 \: \: i.e. \: \: when \: \: \theta = \frac{\pi}{2} \: \: or \: 90$

=> thus, range on maximum height

$\bf = \frac{ {u}^{2} \sin((2 \times \frac{\pi}{2} ) }{g} = \frac{ {u}^{2} \sin(\pi) }{g} = \underline 0$

This indicates that when an object is thrown above it reaches greatest height than when thrown in projectile.
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2)
=> maximum range can be produced when

$\sin(2 \theta ) =1 \: \: \: \: i.e.\\ \\ = > 2 \theta = \frac{\pi}{2} \\ \\ = > \bf \theta = \underline \frac{\pi}{4}$

So, height at maximum range

$= \frac{ {(u \sin( \theta )) }^{2} }{2g} = \frac{ {u}^{2} \sin ^{2} ( \frac{\pi}{4} ) }{2g} \\ \\ = > \frac{ {u}^{2} {( \frac{1}{ \sqrt{2} }) }^{2} }{2g} = \bf \underline{ \frac{ {u}^{2} }{4g} }$
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3)

The height will be maximum at angle 90° or $\frac{\pi}{2}$

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