1 mph 60 mph 4 bob throws a baseball directly up the

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(1) mph = 60 mph. 4. Bob throws a baseball directly up. The ball leaves his hand when it is 2.00 m above the ground. How long does it take for the ball to hit the ground if the maximum height reached by the ball is 4.00 m? (a) 1.54 s 4 (b) 1.28 s (c) 1.81 s (d) 2.02 s (e) 1.01 s ANS: Identify this as a free fall problem. We use y ( t ) = - 1 2 gt 2 + v 0 y t + y 0 , v y ( t ) = - gt + v 0 y and v 2 y - v 2 0 y = - 2 g ( y - y 0 ), with g = 9 . 8 m/s 2 . We choose the y -axis to point up with the origin at ground level, so that y 0 = 2 m. We need to find out what the initial velocity is. The fastest the ball is thrown the higher it will go, so we can figure out the initial velocity from the fact that it goes to 4 m when it is thrown from 2 m. When the ball 4
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reaches maximum height it stops momentarily, so that at that instant, we have v y = 0. Using v 2 y - v 2 0 y = - 2 g ( y - y 0 ), we have 0 - v 2 0 y = - 2 g ( y max - y 0 ), or v 0 y = p 2 g ( y max - y 0 ) = p 2(9 . 8)(4 - 2) m/s = 6 . 26 m/s. We can now use y ( t ) = - 1 2 gt 2 + v 0 y t + y 0 to find the time t g at which the ball hits the ground ( y ( t g ) = 0). Taking t 0 = 0 the time t g is the time it takes the ball to hit the ground. So we solve y ( t g ) = 0, that is 0 = - 1 2 gt 2 g + v 0 y t g + y 0 or 0 = - 1 2 (9 . 8) t 2 g + 6 . 26 t g + 2, which gives two solutions, t g = 1 . 54 s and t g = - 0 . 26 s. We discard the second, unphysical solution. 5. Colin believes that one can save gas by slowly increasing the acceleration from zero so that it reaches 2 mph/s after 1 min. Assuming that this increase is linear in time (that is, a = kt for some constant k ), which graph best represents Colin’s velocity as a function of time, if he started accelerating form rest? (a) 4 (b) (c) (d) (e) ANS: Three ways to approach this: 1. Elimination. The following are wrong: (b) and (d) Constant slope in v vs t means constant acceleration. (d) Constant velocity means vanishing acceleration.
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