Question

The speed of the projectile at maximum height is (1)/(sqrt(2)) times initial speed.If the range of the projectile is N times maximum height attained by it then N equals​

Answer 1

Given : The speed of the projectile at maximum height is 1/√2 times initial speed.

the range of the projectile is N times maximum height attained by it

To Find : Value of  N

Solution:

Let say Initial Velocity V at angle α

Horizontal Speed = VCosα

Vertical Speed = VSinα

The speed of the projectile at maximum height =  VCosα  as Vertical Speed  will be zero

VCosα  =  V/√2

=> Cosα  = 1/√2

=>  α  = 45°

=> VCosα =  VSinα

Let sat t is time for max height then 2t will be time for range

Height

h =  (1/2)( u + v) * t  = (1/2)( VSinα + 0) * t  =  VSinα * t /2

Range = VCosα * (2t)  = 2 VCosαt

VCosα =  VSinα

=> Range =  2  VSinα * t

=> Range = 4 ( VSinα * t /2)

=> Range = 4 h

N = 4

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Answer 2

Given:

The speed of the projectile at maximum height is 1/√2 times initial speed. The range of the projectile is N times maximum height attained.

To find:

Value of N ?

Calculation:

We know that at maximum height, the projectile only has horizontal component.

So, let initial velocity be u :

$\rm \therefore \: u_{max} = \dfrac{1}{ \sqrt{2} } \times u$

$\rm \implies \: u \cos( \theta) = \dfrac{1}{ \sqrt{2} } \times u$

$\rm \implies \: \cos( \theta) = \dfrac{1}{ \sqrt{2} }$

$\rm \implies \: \theta = {45}^{ \circ}$

Now, as per question, the range is N times max height !

$\rm range = N \times( max \: height)$

$\rm \implies \: \dfrac{ {u}^{2} \sin(2 \theta) }{g} = N \times \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

$\rm \implies \: \dfrac{ {u}^{2} 2\sin( \theta) \cos( \theta) }{g} = N \times \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

$\rm \implies \: 2\sin( \theta) \cos( \theta) = N \times \dfrac{ { \sin}^{2}( \theta) }{2}$

$\rm \implies \: \cos( \theta) = N \times \dfrac{ \sin( \theta) }{4}$

$\rm \implies \: \cos( {45}^{ \circ} ) = N \times \dfrac{ \sin( {45}^{ \circ} ) }{4}$

$\rm \implies \: \dfrac{1}{ \sqrt{2} } = N \times \dfrac{ ( \frac{1}{ \sqrt{2} } ) }{4}$

$\rm \implies \: 1 = N \times \dfrac{ 1 }{4}$

$\rm \implies \: N = 4$

So, value of N = 4.