COMPLETE EACH PROBLEM ON A SEPARATE SHEET OF PAPER WITH YOUR NAME ON THE BACK. S

Question

COMPLETE EACH PROBLEM ON A SEPARATE SHEET OF PAPER WITH YOUR NAME ON THE BACK. SHOW YOUR WORK, DRAW A PICTURE, IDENTIFY KNOWN AND UNKNOWN QUANTITIES, SHOW FORMULA, SOLVE ALGEBRAICALLY, AND INCLUDE UNITS WITH ANSWERS TO RECEIVE FULL CREDITl PUT A CIRCLE AROUND YOUR FINAL ANSWER. PROBLEMS (10 points each) 1. A baseball is thrown directly upward (positive y direction) with a velocity of 32.5 m/'s The baseball is 1.70 meters above the ground when it is released. The basehall reaches a maximum height, and then falls back down and hits the ground. a) What is the maximum height above the ground that the baseball will reach? b) How much time will it take to reach the maximum height? c) What is the total amount of time in the air for the baseball? d) What is the displacement of the baseball for the entire motion described? 2. A vector has a magnitude of 47.5 and a direction of 212 degrees with respect to the positive axis. a) Draw a picture of the vector b) Determine the x and y components of the vector o) Write the vector using unit vevctor nsotations Given the vector-4.00/ 7.50) (NOI related to the first pan of the mbhm) d) Draw a picture of the vector e) Find the magnitude and direction of the wwr with est o the tive i

Explanation / Answer

1] a] Maximum height above the ground will be:

H = 1.7 + [v2 - u2]/2g

at the highest point, v = 0 m/s

therefore,

H = 1.7 + [32.52/2g] = 55.590 m

b] v = u - gt

at the maximum height, v = 0 m/s

so, 0 = 32.5 - 9.8t

=> t = 3.316s

c] After reaching the highest point, the time taken to reach the ground can be found using:

H = ut + (1/2)gt2

at the highest point, u = 0 m/s

therefore, 55.590 = 0 + (1/2)(9.8)t2

=> t = 3.368s

therefore, the total time for which the ball is in the air is:

T = 3.316 + 3.368 = 6.6842 s.

d] Displacement of the ball will be: S = (32.52/2g) - 55.590 = - 1.7 m.

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