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larisa86 [58]
3 years ago
14

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 ◦
Physics
1 answer:
allsm [11]3 years ago
8 0

Answer:

\theta_2 \approx 40.5^{\circ}

Explanation:

<u>Given:</u>

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

angle of projection of the first ball, \theta_1= 72^{\circ}

∵The balls should land at the same point,

∴their range of projectile,

R_1=R_2=R\,m

As we know for the range of projectile:

R=\frac{v^2}{g}.sin\,2\theta

∵ we have equal range in both the cases

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

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

\Rightarrow sin 144^{\circ}=sin\,2\theta_2

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

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

\theta_2 \approx 40.5^{\circ}

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