Question

A man is standing on the basketball standard and launching t-shirts out of an air cannon. The t-shirts leave the tube at a height of 425 cm above the basketball floor.The tube consistently launches the t-shirts with an initial velocity of 10.2 m/s. When the tube is angled 33.3° above horizontal how far from the spot on the floor directly below where they are launched will the t-shirts land? (neglect air friction) Finding a symbolic solution and an explanation of how to use that solution will help me understand the problem. (A provided hint is: Find an equation for the time that the t-shirt is in the air before hitting the ground is a good start.)

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Answer 1

3lep! 10-2.5in33 02m/s 632- I0 2 cos33.3 425 cm .4.25m 7 3 tpa usirg ver kical componunt Neloci lo2sin3 my) will nd how n the air we t-shirt ig 2nd eqa o mohon hauttat we -4.25 10 2 sin 33.3. t -9.91xt2 0.Sx9.812-102 sin3s-t-4.2so uadralic era. This is a two valu t - 0.521 X 166 Hi'me (an net be Chooge t 1-66 S 3 eps Now, use horizond Yelbiy to tind out how far tshirt landad. Mind that in horizontal direchon veledy iConutant Re v celo t 10.2 col 33 3x 1.66 Ates 14.15 m

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