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)

a) 10m/s
b) 12m/s
c) 13.42 m/s
d) 15.25 m/s

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

Answer:

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