SOLUTION: The soccer player kicks the ball in the Figure above with an initial velocity of 35 m/s at an angle of 20°. (a) Calculate for the time when it reaches 0.40 m. (b) Find its positio
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Question 1194321: The soccer player kicks the ball in the Figure above with an initial velocity of 35 m/s at an angle of 20°. (a) Calculate for the time when it reaches 0.40 m. (b) Find its position parallel to the field at height equal to 0.40 m. (c) Find the time when it reaches its highest point. (d) At what point is the ball at its highest and farthest?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The soccer player kicks the ball in the Figure above with an initial velocity of 35 m/s at an angle of 20°. (a) Calculate for the time when it reaches 0.40 m. (b) Find its position parallel to the field at height equal to 0.40 m. (c) Find the time when it reaches its highest point. (d) At what point is the ball at its highest and farthest?
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Neglecting air friction and using 9.8m/sec/sec for gravity:
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(a) Calculate for the time when it reaches 0.40 m.
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The vertical component is 35*sin(20) m/sec.
Height h(t) = -4.9t^2 + 35*sin(20)*t = 0.4
4.9t^2 - 35*sin(20)t = -0.4
Solve the quadratic for t. There are 2 values, ascending then descending.
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(b) Find its position parallel to the field at height equal to 0.40 m.
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If that means its distance from where it was kicked:
The horizontal component is 35*cos(20)*t
Calculate using the t values from (a).
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(c) Find the time when it reaches its highest point.
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It's a parabola. The max is at t = -b/2a = 35*sin(20)/9.8
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(d) At what point is the ball at its highest
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At t = the value in (c)
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and farthest?
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Multiply t from (c) times 2 times the horizontal component.
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