Bullet is fired at an angle = 30°
The bullet hits the ground at a distance of 3km = 3000 m
Horizontal range, R = \( \dfrac{ u^2 sin^2θ}{g}\)
3000 =\( \dfrac{u^2 sin 60°}{9.8}\)
u2 = \( \dfrac{ (3000 \times 9.8)}{\dfrac{\sqrt{3}}{2}}\)
= \( \dfrac{ 2 (3000 \times 9.8)}{\sqrt{3}}=\dfrac{58800}{1.732}\) = 33949
Also, R’ = \( \dfrac{ u^2 sin2θ}{g}\) ⇒ 5000
= \( \dfrac{(33949 \times sin2θ)}{9.8}\)
Sin 2θ = \( \dfrac{ (5000 \times 9.8)}{33949 }\)
= \( \dfrac{49000}{33949}\) =1.44
Since sine of an angle cannot be more than 1. Therefore, a target 5 km away cannot be hit
Answered by Pragya Singh | 1 year ago(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by
\(\theta (t)=tan^{-1}(\dfrac{v_{0y-gt}}{v_{ox}})\)
(b) Shows that the projection angle θ0 for a projectile launched from the origin is given by
\(\theta_{0}=tan^{-1}(\dfrac{4h_{m}}{R})\)
where the symbols have their usual meaning
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