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Question:

What must be the initial velocity of an arrow aimed at 45° above the horizontal if it's target is 2.0 m away?

Choose the letter of the correct answer:

A. 2.2m/s
B. 4.4m/s
C. 6.6m/s
D. 8.8m/s
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Sagot :

PASAGOTASAPIN

Answer:

= V^2 Sin2θ / g

In this case, R = 2.0m, θ = 30°, so

2.0 = V^2(0.866)/9.81 then V = √(2.0x9.81)/0.866 = 4.76m/s

Explanation:

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