Question

The minimum speed of projection required to strike the top of a pole of height 8m and located at a horizontal distance of 6m from the point of projection is nearly (g= 10m/s^(2))​

Answer 1

Given : projection required to strike the top of a pole of height 8m and located at a horizontal distance of 6m from the point of projection

To Find : Minimum Speed to strike the top of a pole

a) 10m/s

b) 12m/s

c) 13.42 m/s

d) 15.25 m/s

Solution:

Let say velocity = V at angle α

T is the the time of strike

VCosα.T = 6

=> VCosα. = 6/T

S = ut + (1/2)at²

t = T , u =  VSinα  ,a  = - g = -10

8  = VSinα.T + (1/2)(-g)T²

=> 8 = VSinα.T - (1/2)(10)T²

=> 8 = VSinα.T - 5T²

=> VSinα.T = 8 + 5T²

=> VSinα = 8/T + 5T

V²Cos²α + V²Sin²α = V²   ∵  Cos²α +  Sin²α =

=>  V² = (6/T)² + (8/T + 5T)²

V²  will be minimum when V is minimum

 V² = Z

=> Z = (6/T)² + (8/T + 5T)²

=> Z  = 36/T²  + 64/T²  + 25T²  + 80

=> Z = 100/T² + 25T² + 80

dZ/dT = -200/T³  + 50T  

-200/T³  + 50T = 0

=> T⁴  = 4

=> T² = 2

=>  T  = √2

d²Z/dT² = 600/T⁴  + 50 > 0

Hence minimum at T² = 2 or   T  = √2

V² =  100/T² + 25T² + 80

=> V² =  50 + 50 + 80

=>  V² =  180

=> V = 13.416  m/s

=> V = 13.42  m/s

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Answer 2

To find:

The minimum speed of projection required to strike the top of a pole of height 8m and located at a horizontal distance of 6m from the point of projection is ?

Calculation:

The general expression for trajectory of Projectile is :

$\boxed{ \bold{ \therefore \: y = x \tan( \theta) - \dfrac{g {x}^{2} }{2 {u}^{2} { \cos}^{2} ( \theta)} }}$

Now , since object passes through coordinate (6,8) , we can put those values:

$\implies \: 8 = 6 \tan( \theta) - \dfrac{g {(6)}^{2} }{2 {u}^{2} { \cos}^{2} ( \theta)}$

$\implies \: 8 = 6 \tan( \theta) - \dfrac{10 \times 36}{2 {u}^{2} { \cos}^{2} ( \theta)}$

$\implies \: 8 = 6 \tan( \theta) - \dfrac{180}{ {u}^{2} { \cos}^{2} ( \theta)}$

$\implies \: 4 = 3 \tan( \theta) - \dfrac{90}{ {u}^{2} { \cos}^{2} ( \theta)}$

$\implies \: \dfrac{90}{ {u}^{2} { \cos}^{2} ( \theta)} = 3 \tan( \theta) - 4$

$\implies \: \dfrac{90 \: { \sec}^{2} ( \theta)}{ {u}^{2}} = 3 \tan( \theta) - 4$

$\implies \: \dfrac{90 \: \{1 + { \tan}^{2} ( \theta) \}}{ {u}^{2}} = 3 \tan( \theta) - 4$

$\implies \: {u}^{2} = \dfrac{90 \: \{1 + { \tan}^{2} ( \theta) \}}{3 \tan( \theta) - 4}$

Now , if u has minimum value , then u² will also have minimum value :

$\implies \: Z = \dfrac{90 \: \{1 + { \tan}^{2} ( \theta) \}}{3 \tan( \theta) - 4}$

For minima , $\dfrac{dZ}{d\theta}$ should be zero and $\dfrac{d^{2}Z}{d\theta^{2}}$ should be > 0 :

$\implies \: \dfrac{dZ}{d \theta} = 0$

$\implies \: 3 { \sec }^{2} ( \theta) \{1 + { \tan}^{2}( \theta) \} - \{3 \tan( \theta) - 4 \} \{2 \tan( \theta) { \sec}^{2}( \theta) \} = 0$

$\implies \: 3 \{1 + { \tan}^{2}( \theta) \} - \{3 \tan( \theta) - 4 \} \{2 \tan( \theta) \} = 0$

$\implies \: 3 { \tan}^{2}( \theta) - 8 \tan( \theta) - 3 = 0$

$\implies \: \bigg \{3 \tan( \theta) + 1 \bigg \} \bigg \{ \tan( \theta) - 3 \bigg \}= 0$

Either $\tan(\theta) =- \dfrac{1}{3} \:or \: 3$

After 2nd order differentiation (I haven't shown the 2° differentiation) , we will see that $\dfrac{d^{2}Z}{d\theta^{2}}$ > 0 , when $\tan(\theta)=3$.

Now continuing with the 1st equation:

$\implies \: {u}^{2} = \dfrac{90 \: \{1 + { \tan}^{2} ( \theta) \}}{3 \tan( \theta) - 4}$

$\implies \: {u}^{2} = \dfrac{90 \: \{1 + { 3}^{2} \}}{3( 3)- 4}$

$\implies \: {u}^{2} = \dfrac{90 \: \{1 + 9\}}{5}$

$\implies \: {u}^{2} = 180$

$\implies \: u \approx \: 13.41 \: m {s}^{ - 1}$

So, minimum velocity with which the Projectile can pass through coordinate (6,8) is 13.41 m/s.

• My method is much more difficult than the previous answer.

• I would recommend you to follow the previous answer by Amitnrw sir and use that method in subjective exams.