A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R=
.If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig), show that it could be achieved by raising the gun to a height at least h =
A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R= .If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig), show that it could be achieved by raising the gun to a height at least h =
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1 Answer
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This is a Long Answer Type Question as classified in NCERT Exemplar
Explanation – target T is at horizontal distance x= R+ and between point of projection y= -h
Maximum horizontal range R= …………1
Horizontal component of initial velocity = Vocos
Vertical component of initial velocity = -Vosin
So h = (-Vosin )t + 2………….2
R+ = Vocos
So t=
Substituting value of t in 2 we get
So h = (-V0sin )
H = -(R+ )tan +
, h = -(R+ )tan +
So h = -(R+ ) +
So h = -(R+ )+
So h = -R- +(R+ )
h=
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Please find the solution below:
after 10 kicks,
v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec
ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²
a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r
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