Question
Asked Sep 18, 2019
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A shot-putter throws the "shot" (mass = 7.3 kg) with an initial speed of 14.4 m/s at a 34.0o angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 21.0m above the ground. 

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Expert Answer

Step 1

The vertical and horizontal velocity of the shot-putter is,

Find the time interval for the shot is travelled in vertical motion, by using displacement.

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sine at y=yo+f+ 2 0 h+vosin 2

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Step 2

Solve the above quadratic equation for time.

Substitute the values,

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V0 sin 0tvsin20+2gh -g -(14.4)sin34.0°t y(14.4) sin'(34.0°)+2(98)(21.0) -(98) t= 8.05-05.96 9.8 -8.05 -10.29 9.8 =1.87s

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Step 3

The horizontal distance travelled by the sho...

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L=v fcos L-(14.4m/s) (1.87s)cos(34.0°) 22.3m

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Kinematics

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