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
Answers
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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