Kevin S. Huang

2007: Problem 24

Topic: Kinematics
Concepts: Projectile motion, Vectors

Solution:

There is a constant net force $F_{\text{net}}$ acting on the ball so it has to be thrown opposite to $F_{\text{net}}$ to come back to its starting position. We have

\tan\theta=\frac{mg}{F_0}

\theta=\arctan\left(\frac{mg}{F_0}\right)

so the answer is B.

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kshc027

2007, F=ma Exam

Fma: Kinematics, Projectile motion, Vectors

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