A major leaguer hits a baseball so that it leaves the bat at a speed of 39.0 m/s
ID: 1590703 • Letter: A
Question
A major leaguer hits a baseball so that it leaves the bat at a speed of 39.0 m/s and at an angle of 36.9° above the horizontal. You can ignore air resistance. (Assume upward is positive.)
(a) At what two times is the baseball at a height of 13.0 m above the point at which it left the bat?
first time ?
second time ?
(b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a).
first time ?
in (horizontal component)
in (vertical component)
second time ?
in (horizontal component)
in (vertical component)
(c) What are the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?
m/s ?
(below the horizontal) ? in degrees
Explanation / Answer
a) Here, 13 = 39*sin36.9*t - 0.5 *9.8*t2
=> 13 = 23.41t - 4.9t2
=> 4.9t2 - 23.41t + 13 =0
=> first time ,t1 = 0.6414 sec
second time , t2 = 4.136 sec
b) For first time t1 , vertical velocity = 39*sin36.9 - 9.8*0.6414
= 17.13 m/sec -----> vertically upwards
horizontal velocity = 39*cos36.9
= 31.19 m/sec ---------> towards right
For second time t2 , vertical velocity = 39*sin36.9 - 9.8*4.136
= - 17.13 m/sec ----------> vertically downwards
horizontal velocity = 39*cos36.9
= 31.19 m/sec -----------> towards right
c) magnitude of ball velocity when it returns to the level at which it left the bat = 39 m/sec
Direction = - 36.9 degrees below horizontal .
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