iPad 0:14 AM *100% session.masteringphysics.com Anindya Sengupt...Wednesday\'s R
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iPad 0:14 AM *100% session.masteringphysics.com Anindya Sengupt...Wednesday's Rar. Pearson's MyLab.MasteringPhysics MasteringPhys tamu BIMS 289 5 Week 2 (graded) Problem 3.48 19 of20 > Constants A baseball thrown at an angle of 65.0 above the horizontal strikes a building 18.0 m away at a point 7.00 m above the point from which it is theown. ignore air resistance Part A Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown. 19.5 m/s Submit Incorrect: Try Again; 3 attempts remaining Part B Find the magnitude ofthe velocity of the baseball just before stries the buldng. m/s Submit Request Answer PartC Find the direction of the velocity of the baseball just belore it sorkes the building. below the horizontal Submit Provide Feedback Next >Explanation / Answer
a)
The vertical component of initial velocity is u sin = u sin 65
= 0.91u.
The time to have a vertical displacement of 7 m is given by
s = ut + 1/2at^2
7 = 0.91u t - 4.9 t^2-------------------1
Horizontal distance traveled is 18 m = u cos 65*t. = 0.42 u t.
u t = 42.86
Squaring both sides
[ut]^2 = 1836.7
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Eliminating t and t^2 from 1
7 = 39 - 9000 /u^2
u = 16.8 m/s
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b)
Using conservation of energy at initial point and final point
v^2-u^2 = 2as,
v^2 = 16.78^2 - 2*9.8*7
v = - 12 m/s minus to indicate that it points down.
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c)
The horizontal component of 12.01 is 12.01cos = 16.78cos65
From which = 54
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