hansen44

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A ball is thrown vertically with a speed of 3 m/s. Find the total time the ball has been in flight when it has traveled 50 cm.

I cannot get the right answer, I mean this a simple kinematic question but I keep getting the wrong answer. Can someone please do the problem and tell me what you get and how you did it. I would really appreciate it. Thanks
 

BloodySurgeon

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A ball is thrown vertically with a speed of 3 m/s. Find the total time the ball has been in flight when it has traveled 50 cm.

I cannot get the right answer, I mean this a simple kinematic question but I keep getting the wrong answer. Can someone please do the problem and tell me what you get and how you did it. I would really appreciate it. Thanks
Unless the maximum height is 50cm (which is not), I don't think the mcat will ask this question because it involves a quadratic equation. And if it is the maximum height than the mcat will tell you this or do so implicitly. However, you can still estimate the answer as the time will be slightly longer than reaching the apex but less than twice that time.
 

PRodulous

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If 50 cm is the apex (which i don't think it is), is the answer around 2.2 seconds?

Two things. First, I revised my work and it seems that this is a somewhat long problem with about 4 steps that may not be on the MCAT. However, it's so complex that it has made a simple kinetics problem into a brain exercise.

I've changed my answer to 0.4 seconds. I can post my work if you like.
 
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hansen44

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Yes the answer is t=.398 but how did you know the apex was 45 cm? Exactly how did you figure it out Bloody Surgeon?
 

PRodulous

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Yes the answer is t=.398 but how did you know the apex was 45 cm? Exactly how did you figure it out Bloody Surgeon?
To find apex:
v=vot + at^2 (at apex, v=0)
0=(3)t + at^2 (t factored out)
0=3 + at
-(-10)t=3
t= 3/10 seconds to apex

distance to apex:
d=vot + (1/2)at^2
d=(3)(3/10) + (1/2)(-10)(3/10)^2
d=(9/10) - (9/20) = 9/20 = 45/100 = 0.45 meters or 45 cm

time to travel additional 0.05 meters
d=vot + (1/2)at^2
0.05=0(t) + (1/2)(-10)t^2
0.05 = (1/2)(10)t^2
(0.05)(2) = 0.01 = t^2
t^2 = 0.01 = 1/100
t=1/10 = 0.05 seconds

Add times together:
t total = (3/10) + (1/10) = 4/10 or 0.4 seconds
 

BloodySurgeon

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This is the easy way:

Vf^2 = Vi^2 + 2ah
(3)^2 = (20)h
h = 9/20 or 4.5/10 = 45cm

Vf = Vi + at
3=10t
t=3/10 or 0.3s

total time (round trip) = 2(0.3)= 0.6

so answer is around 0.3 and 0.6 but closer to 0.3 so I guessed 0.4
 
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hansen44

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Ok thanks everybody for your help I see now. I have another question thats bothering me.

If F(drag)=bv^2 for a ball of mass 3kg, and the terminal velocity of the ball is 30 m/s, what is b.

If you tell me roughly how you did this I would appreciate it, I have these physics problems but no solutions only answers so its difficult to see exactly how they solved it. thanks
 

PRodulous

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Ok thanks everybody for your help I see now. I have another question thats bothering me.

If F(drag)=bv^2 for a ball of mass 3kg, and the terminal velocity of the ball is 30 m/s, what is b.

If you tell me roughly how you did this I would appreciate it, I have these physics problems but no solutions only answers so its difficult to see exactly how they solved it. thanks
At terminal velocity, drag force=force of gravity. So:
mg=bv^2
b=(mg)/(v^2)=(3*10)/(30^2)=30/900=1/30 kg/s
 
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