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01 Sep 2020, 01:20
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61% (02:33) correct
39%
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The velocity of a propelled object is described by the equation v=v0+at, where v0 is the object's initial velocity, a is acceleration, and t is time. The distance that the object travels, d, is given by v0t+ 1/2 a\(t^2\). If a ball is launched from the ground straight up into the air at an initial velocity of 19.6 meters per second, and the acceleration due to gravity is −9.8 meters per second squared, how high will the ball travel before it begins to fall back down?
A. 19.6 meters
B. 28.8 meters
C. 39.2 meters
D. 58.8 meters
E. 78.4 meters
A. 19.6 meters
B. 28.8 meters
C. 39.2 meters
D. 58.8 meters
E. 78.4 meters
01 Sep 2020, 01:43
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The maximum height will be reached when the ball stops traveling upward and is about to begin its downward flight, as indicated by the wording "How high will the ball travel before it begins to fall back down." At the instant where the ball has traveled as high as it will go, and has not yet begun to fall back down, its' velocity is zero, so v = 0.
Plug that into the equation for velocity to get the value for t when that occurs: 0 = 19.6 − 9.8t.
Thus, 9.8t = 19.6 and t = 2. Now that you have the value of t, you can plug it into the equation for distance:
d=v0t+1/2*a \(t^2\)=19.6(2)+1/2(−9.8)(2)(2)
This simplifies to 39.2 − 19.6 = 19.6 meters.
Hence (A) is correct.
Encourage the solution with Kudos... Thanks.
Plug that into the equation for velocity to get the value for t when that occurs: 0 = 19.6 − 9.8t.
Thus, 9.8t = 19.6 and t = 2. Now that you have the value of t, you can plug it into the equation for distance:
d=v0t+1/2*a \(t^2\)=19.6(2)+1/2(−9.8)(2)(2)
This simplifies to 39.2 − 19.6 = 19.6 meters.
Hence (A) is correct.
Encourage the solution with Kudos... Thanks.
General Discussion
01 Sep 2020, 11:19
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To answer this question, one has to understand the science behind gravity (v = 0 just before falling down on the ground) can such questions really come on gmat??
