A person catches a football 5.2 seconds after it is thrown. If the average acceleration of the football is -1.2 m/s2, what is the velocity of the ball when it leaves the hand of the thrower? (Assume that “towards the receiver” is positive.)

Question

18jonea · Answer

@mathmale

mathmale · Answer

If the acceleration, a, is constant, then $$v=v _{0}+a*t$$

mathmale · Answer

It would seem to me that v=0 when the ball is caught.  Since you are given a and t, you should be able to work backwards and find the initial velocity.  Try it.  Does your answer seem to make sense?

18jonea · Answer

these are my choices 	
6.4 m/s
 	
4.0 m/s
 	
6.2 m/s
 	
-4.3 m/s

18jonea · Answer

@mathmale

mathmale · Answer

I have given you a formula appropriate to solving this problem, and have reminded you that the acceleration, a=-1.2 m/(sec)^2 and the elapsed time 5.2 sec.  You need to insert these values into the formula$$v=v _{0}+a*t$$   Once you have a result, compare your result to the four answer choices.

18jonea · Answer

i got 6.24 would that be correct? @mathmale

18jonea · Answer

@mathmale

mathmale · Answer

This problem may be harder than it appears.  I get the same result when multiplying -1.2 by 5.2:  -6.24 m/sec.

But this assumes that the velocity of the ball is zero when caught.  That would be very rare indeed!  So, all we can really say is that the ball's velocity became less and less (-6.4 m/sec) as it traveled from the thrower to the catcher.  I don't see how we can calculate the initial velocity if we don't know how fast the ball is moving when caught.

So, again, the DIFFERENCE in velocity, start to finish, is -6.4 m/sec.