Question

the minimum speed of projection required to strike the top of a pole of height 8 metre and located at a horizontal distance of 6 metre from the point half projection is nearly (g=10m/s²)​

Answer 1

Given:

Height of the pole = 8 m

Horizontal distance of the pole from the point of projection = 6 m

g = 10²

To find:

Minimum speed of projection required to strike the top of the pole.

Solution:

Let us assume the velocity = V

And the angle of projection be α

Let the time of strike  = T

As we know that the horizontal distance is 6 m

Vcosα*T = 6

Vcosα = 6/T

As we know that:

S = ut + (1/2)at²

Putting the values we get:  

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   because  cos²α +  sin²α =  1

 V² = (6/T)² + (8/T + 5T)²

V will be minimum when V² would be minimum.

 V² = (6/T)² + (8/T + 5T)²

V² = 36/T²  + 64/T²  + 25T²  + 80

V²= 100/T² + 25T² + 80

dV²/dT = -200/T³  + 50T  

Putting dV²/dt = 0

-200/T³  + 50T = 0

T⁴  = 4

 T  = √2 , at this time V will be minimum since d²V/dT² = 600/T⁴  + 50  which is greater than 0

V² =  100/T² + 25T² + 80

V² =  50 + 50 + 80

V² =  180

V = 13.416  m/s

V = 13.42  m/s

Therefore the minimum velocity required is 13.42 m/s.