Physics motion webassign

1. Sep 9, 2010

fparks

1. The problem statement, all variables and given/known data
A tennis ball is thrown straight up with an initial speed of 25.0 m/s. It is caught at the same distance above the ground.

(a) How high does the ball rise?
_________m

(b) How long does the ball remain in the air? (Hint: The time it takes the ball to rise equals the time it takes to fall.)
_________s

2. Relevant equations

3. The attempt at a solution
i have no idea how to do this. please help!

2. Nov 14, 2010

ONIphysics

here's your first equation:
T-time Vf-final velocity
Vi-initial velocity A-acceleration
T=(Vf-Vi)/A
the object is going up, so all Velocity is up (no horizontal motion)
vertical acceleration is always -9.8 (due to gravity)
Vi=25m/s, and Vf is 0 at the top (where max vertical distance is reached) so,
T=(0-25)/-9.8 OR T=25/9.8 <(T for max height)
simple multiply by two, and you have your answer for number 3.

your second equation would be:
Dy-change in y distance (vertical), T, Vi, A, and angle (90)
Dy=Vi(sin(90))t+1/2at^2 so,
Dy=25(1)t-4.9t^2 OR Dy=25(25/9.8)-4.9(25/9.8)^2

therefore, T~5.102 <(total T) and Dy~31.888m
where ~ means approximately equal to

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