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A high jumper leaves the ground with an initial velocity of 10 m/s rightward at an angle of 15°. What is the high jumper's initial horizontal velocity?

a. 10 m/s
b. 2.5 m/s
c. 9.6 m/s
d. 10 cos(15°) m/s

User Razimbres
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1 Answer

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Final answer:

The initial horizontal velocity of a high jumper with an initial velocity of 10 m/s at an angle of 15° is calculated using the cosine component of the initial velocity, yielding the answer d. 10 cos(15°) m/s.

Step-by-step explanation:

The student is asked to calculate the initial horizontal velocity of a high jumper who leaves the ground with an initial velocity of 10 m/s at an angle of 15°. The initial horizontal velocity can be calculated using the cosine component of the initial velocity vector. The formula to find the horizontal component (vx) is vx = v * cos(θ), where v is the initial velocity and θ is the angle of projection relative to the horizontal.

Now, applying the formula to the given values:

vx = 10 m/s * cos(15°)

Upon calculating, we find that the correct answer is d. 10 cos(15°) m/s, which will provide the high jumper's initial horizontal velocity.

User Bogdan Osyka
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