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svlad2 [7]
3 years ago
12

A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

n of magnitude a opposite the direction of motion. a. Find an expression in terms of a and g for the launch angle that gives maximum range. b. What is the angle for maximum range if a is 10% of g?
Physics
1 answer:
ale4655 [162]3 years ago
7 0

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is v_0

Horizontal acceleration is a_x=a

At maximum height velocity is zero therefore

v_f=v_i-gt

0=v_0\sin u-gt

t=\frac{v_0\sin u}{g}

Total time of flight T=2t=\frac{2v_0\sin u}{g}

During this time horizontal range is

R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}

R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}

For maximum range \frac{\mathrm{d} R}{\mathrm{d} u}=0

\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}

\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0

\tan 2u=\frac{g}{a}

u=\frac{1}{2}tan ^{-1}\frac{g}{a}

(b)If a =10% g

a=0.1g

thus u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}

u=42.14^{\circ}

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