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One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching this first snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 16.1 m/s. The first one is thrown at an angle of 72◦ with respect to the horizontal. At what angle should the second snowball be thrown to arrive at the same point as the first? Answer in units of ◦

Answer :

Answer:

[tex]\theta_2 \approx 40.5^{\circ}[/tex]

Explanation:

Given:

velocity of the ball first and ball second, [tex]v_1=v_2=v\,m.s^{-1}[/tex]

angle of projection of the first ball, [tex]\theta_1= 72^{\circ}[/tex]

∵The balls should land at the same point,

∴their range of projectile,

[tex]R_1=R_2=R\,m[/tex]

As we know for the range of projectile:

[tex]R=\frac{v^2}{g}.sin\,2\theta[/tex]

∵ we have equal range in both the cases

[tex]\therefore \frac{v_1^2}{g}.sin\,2\theta_1=\frac{v_2^2}{g}.sin\,2\theta_2[/tex]

[tex]\Rightarrow \frac{v^2}{g}.sin\,(2\times 72) =\frac{v^2}{g}.sin\,2\theta_2[/tex]

[tex]\Rightarrow sin 144^{\circ}=sin\,2\theta_2[/tex]

[tex]2\theta_2=sin^{-1}[sin 144^{\circ}][/tex]

[tex]\theta_2=\frac{1}{2}\times sin^{-1}[sin 144^{\circ}][/tex]

[tex]\theta_2 \approx 40.5^{\circ}[/tex]

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