Suppose we throw a baseball with an initial speed of $12 \mathrm{~m} / \mathrm{s}$ in a direction $60^{\circ}$ upward from the horizontal. What is the baseball's speed at the peak of its trajectory? (Hint: You do not need to do a lot of calculating here.)
A) $12 \mathrm{~m} / \mathrm{s}$
B) $10.4 \mathrm{~m} / \mathrm{s}$
C) $6 \mathrm{~m} / \mathrm{s}$
D) $3 \mathrm{~m} / \mathrm{s}$
E) $0 \mathrm{~m} / \mathrm{s}$
F) Other (specify)

Question

Quizplus · Accepted Answer

At the peak of its trajectory, the vertical component of the baseball's velocity is 0 m/s due to gravity, but the horizontal component remains unchanged. The initial speed of 12 m/s at a $60^{\circ}$ angle means the horizontal component is $12 \cos(60^{\circ}) = 6 \mathrm{~m} / \mathrm{s}$, which is the speed at the peak.

Suppose We Throw a Baseball with an Initial Speed Of 12