PHY 121
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A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
How high does it rise and how long does it take to get to its highest point?
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
At what clock time(s) will the speed of the ball be 5 meters / second?
At what clock time(s) will the ball be 20 meters above the ground?
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
Question 1a. y = (v^2 v0^2 / 2 a) + 12 m
y = ((0 (15 m/s)^2) / 2 ( -10 m/s^2)) +12 m
y = ((0 225 m/s) / -20 m/s^2) + 12 m
y = (-225 m/s / -20 m/s^2) + 12 m
y = 11.25 m + 12 m
y = 23.25 m is how high the ball goes in the air.
Question 1b. y = -( v0 / a)
y = - (15 m/s / (-10 m/s^2))
y = -(-1.5 s)
y = 1.5 s is how long it takes it to reach its highest point.
Question 2a. y = v0t + ½ at^2
0 = (15 m/s)t + ½ (-10 m/s^2)t^2
0 = (15 m/s 5 m/s^2t)t
t = 15 m/s / 5 m/s^2
t = 3 s is how long it take for the ball to hit the ground.
Question 2b. v = v0 + at
v = 15 m/s (10 m/s^2) (3 s)
v = 15 m/s 30 m/s
v = -15 m/s is the velocity of the ball when it hits the ground.
Question 3. v = v0 + at
v = 15 m/s (10 m/s^2)( 5 s) = -35 m/s
Question 4. y = y0 + v0t + ½ at^2
20 m = 0 + (15 m/s)t + ½ (-10 m/s^2)t^2
(-5 m/s^2)t^2 (15 m/s)t + (20 m) = 0
t = 15 m/s +-sqrt((15 m/s)^2 4(-5 m/s^2)(20 m)) / 2(5 m/s^2)
t = 15 m/s +-sqrt(225 m/s (-400 m/s)) / 10 m/s^2
t = 15 m/s +-sqrt(625) / 10 m/s^2
t = (15 m/s +- 25) / 10 m/s^2
t = (15 m/s + 25) / 10 m/s^2 = 4 s
t = (15 m/s 25) / 10 m/s^2 = -1 s
Question 5. If it hits the ground after 3 s it would be at 0 after 6 seconds.
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30 minutes
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Good responses. Let me know if you have questions.