- #1
hytuoc
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Plz show me how to do #1? and please check my work on #2 and 3. thanks so much! 
1) You drive on Interstate 10 from San Antonio to Houston, half the time at 55km/h and the other half at 90km/h. On the way back you travel half the distance at 55km/h and the other half at 90km/h. What is ur averge speed rom Houston back to San Antonio?
Plz check my work below to see whether if I did it right?
2) A hoodlum throws a stone vertically downward w/ an initial speed of 12.0 m/s from the roof of a building, 30.0 m above the ground. (a) How long does it take the stone to reach the ground ?
(b) What is the speed of the stone at impact?
a) x = xi +vi t + 1/2 at^2
0= 30m + 12m/s t -1/2 * 9.8*t^2
t = 3.99 s or 4.0 s
b) vf = vi + at
= 12 - 9.8 (4s) = -27.2 m/s
****did I do (b) correctly?
3) A basketball player, standing near the basket to grab a rebound, jumps 76.0 cm vertically. How much (total) time does the player spend (a) in the top 15.0 cm of this jump and (b) in the bottom 15.0 cm? Does this help explain y such players seem to hang in the air at the tops of their jumps?
a) xf = xi +vi*t+1/2 at^2
76 = 0 + 0 + 1/2 (9.8) t ^2
t = 3.94 s
61 = 0 + 0 + 1/2 (9.8) t^2
t = 3.53 s
=> t= .41s <- time the player spent in the top 15.0 cm
b) 15 = 0 + 0 + 1/2 (9.8) t^2
t = 1.75 s
1) You drive on Interstate 10 from San Antonio to Houston, half the time at 55km/h and the other half at 90km/h. On the way back you travel half the distance at 55km/h and the other half at 90km/h. What is ur averge speed rom Houston back to San Antonio?
Plz check my work below to see whether if I did it right?
2) A hoodlum throws a stone vertically downward w/ an initial speed of 12.0 m/s from the roof of a building, 30.0 m above the ground. (a) How long does it take the stone to reach the ground ?
(b) What is the speed of the stone at impact?
a) x = xi +vi t + 1/2 at^2
0= 30m + 12m/s t -1/2 * 9.8*t^2
t = 3.99 s or 4.0 s
b) vf = vi + at
= 12 - 9.8 (4s) = -27.2 m/s
****did I do (b) correctly?
3) A basketball player, standing near the basket to grab a rebound, jumps 76.0 cm vertically. How much (total) time does the player spend (a) in the top 15.0 cm of this jump and (b) in the bottom 15.0 cm? Does this help explain y such players seem to hang in the air at the tops of their jumps?
a) xf = xi +vi*t+1/2 at^2
76 = 0 + 0 + 1/2 (9.8) t ^2
t = 3.94 s
61 = 0 + 0 + 1/2 (9.8) t^2
t = 3.53 s
=> t= .41s <- time the player spent in the top 15.0 cm
b) 15 = 0 + 0 + 1/2 (9.8) t^2
t = 1.75 s