
#1
Aug2808, 12:24 AM

P: 83

1. The problem statement, all variables and given/known data
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 15 m/s when the hand is 2.0 m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.) 2. Relevant equations X= Vt + .5at^2 3. The attempt at a solution I'm not sure why I can't get this right. At first I thought it impossible without the acceleration, but then I discovered that because the acceleration is merely the slope of the velocity graph, it would just equal '2/t', right? I have my X (delta x) as 2. (Would it be negative 2??) My initial velocity is 15 m/s (as stated) and my variable is t. I ended with this equation: 2= 15t + .5(2/t)t^2 Why can I not get the answer from this equation? I keep getting extremely low decimals. 



#2
Aug2808, 01:21 AM

Sci Advisor
HW Helper
P: 1,276

Usually in these sorts of questions, you can assume the student is on the surface of the earth, and thus you can use the acceleration due to gravity as the acceleration (ie. a = g = 9.8m/s^2)




#3
Aug2808, 01:25 AM

P: 37

If this problem is occurring on earth, then the acceleration is the acceleration due to gravity which to 2 s.f. is 9.8ms2.
If you define upwards as positive then you must use 9.8 ms2 since gravity is downwards. Also, the initial vertical position was +2m from the ground, while the final is 0m. Thus the final change in displacement (X) was 2 (since it went downwards from the start). Try subbing these values in. And also, the slope of the velocity graph is not 2/t. If you were to calculate the slope using the total time of flight (t), you would need the initial and final velocities. Then you slope would be rise / run = final  initial velocity / time. But even if you tried you would still end up using the actual value of g = 9.8 anyway. 



#4
Aug2808, 01:28 AM

Sci Advisor
P: 1,724

A Student throws a ball straight up... 



#5
Aug2808, 10:06 AM

HW Helper
P: 5,346

Second of all you might consider solving the problem in 2 parts. First the time to max height which since earth gravity is constant is merely t = V_{o}/ g  the constant. Then with the time you can figure X_{max} from the relationship x = 1/2 gt^{2} Then you can figure the return to the ground which is the max height from that plus the additional 2 meters. Putting the new x back in the same equation yields the return time. Answer = time to max + time to ground. 


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