Question
Asked Dec 17, 2019
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A football receiver running straight downfield at 5.50 m/s
is 10.0 m in front of the quarterback when a pass is thrown
downfield at 25.0° above the horizon (Fig. P3.58). If the
receiver never changes speed and the ball is caught at the
same height from which it was thrown, find (a) the football's
initial speed, (b) the amount of time the football spends in
the air, and (c) the distance between the quarterback and the
receiver when the catch is made.
5.50 m/s
-10.0 m-
Figure P3.58
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A football receiver running straight downfield at 5.50 m/s is 10.0 m in front of the quarterback when a pass is thrown downfield at 25.0° above the horizon (Fig. P3.58). If the receiver never changes speed and the ball is caught at the same height from which it was thrown, find (a) the football's initial speed, (b) the amount of time the football spends in the air, and (c) the distance between the quarterback and the receiver when the catch is made. 5.50 m/s -10.0 m- Figure P3.58

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Expert Answer

Step 1

Given information:

Speed of the receiver (vr) = 5.50 m/s

Angle at which the ball is thrown (θ) = 250

The distance of the receiver from the throw (d) = 10.00 m

Step 2

Let “T” be the time spent by the ball in air, then the horizontal distance travelled by the ball in this time is given by:

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u? sin 2(25) = 0.078 u? Where the "" is the velocity of the ball thrown, The distance from the throw at which the receiver catches the ball is given by: 2u sine 10.0 + 5.50 ( :10.0 + 0.473u Equating both the equations, we get: 0.078u? = 10.0 + 0.473u → 0.078u? – 0.473u – 10.0 = 0 Solving the equation gives the velocity as: u = 14.75 m/s

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Step 3

The total time of flight is g...

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2u sin (25) т 2(14.75) sin(25) 1.27 s 9.81 The distance between the quarterback and the receiver after the catch is made, is given by: s = 10.0 + 5.50(T) = 10.0 + 5.50(1.27) = 16.98 = 17 m

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