Question 8 anew hame acid , aoPU n (11 ) Kevin McCarthy midterm1-form210.pdf Sto

Question

Question 8

anew hame acid , aoPU n (11 ) Kevin McCarthy midterm1-form210.pdf Stony Brook University tUS) https:/blackboard.stonybrook edu/bbecswebdav/pid-780411-dt-announcement-rid-28 290669_1/courses/1178-p PHYISI Midterm -Extra, Exam8 888888888 (form: 216) 6. An speed after it has fallen 6.00 m in the vertical direction. object is shot horizontally with a speed of v 16.0 ms off of a very tall cliff. Find the A. 102 ms- B. 138 ms C. 26.8 ms-1 D. 19.3 m-s-1 E. None of the above. same object as in the previous question. How far has the object (initially throwrn horizontally with t 16.0 m-s-1) fallen after 1.21 s? A. -19.4 m C.-0.375 m D.-0.732 m 8. Consider the same object as horizontally with tr = 16.0 m s-1) going after horizontal direction. in the two previous questions. What direction is the object (thrown 1.21 s? The angle is measured relative to the A. 4.32 o (upward) B. -36.5 (downward) C. -20.3 (downward) D. -24.2 . (downward) E. -53.5 (downward) 9. Have you signed side two of the answer sheet over the words to acknowledge that your work on this exam has been done in accordance with an expectation of academic honesty? " Generál Purpose Answer Sheet" 3 5 6 7 8 9

Explanation / Answer

8)(B)-36.5 deg downward

explanation

After time t = 1.21 s, the vertical velocity of the freely faling object is:

vy = g t = 9.81 x 1.21 = 11.87 m/s

while the horizontal velocity remains vx = 16 m/s

so the angle at t = 1.21 s will be:

theta = tan^-1(11.87/16) = 36.5 deg

Hence, (B)-36.5 deg (downward)

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