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A football player kicks the ball with an initial velocity of 28.3 m/s at an angle of 60° above the horizontal. The ball is in the air for a total of 5.00 s before hitting the ground. Ignoring air resistance, what would be the ball's horizontal displacement?

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leena

Hi there!

Since the ball is directed 60° above the horizontal, we must use its HORIZONTAL component to find its displacement.

We can take the cosine to do so:

vcosθ = 28.3cos(60) = 14.15 m/s

Since the ball is in the air for 5 sec, we can use the following equation:

displacement = velocity · time (d = vt)

Plug in the values:

d = 14.15 · 5 = 70.75 m

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