What Are the Peak Height and Total Airtime of a Tennis Ball Thrown Vertically?

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A tennis ball thrown vertically with an initial speed of 25.0 m/s reaches a peak height of approximately 31.888 meters. The time it remains in the air is about 5.102 seconds, as the ascent and descent times are equal. The calculations involve using the equations of motion, factoring in gravitational acceleration of -9.8 m/s². The final velocity at the peak height is zero, allowing for straightforward application of kinematic equations. This analysis effectively demonstrates the principles of projectile motion in a vertical context.
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Homework Statement


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


Homework Equations





The Attempt at a Solution


i have no idea how to do this. please help!
 
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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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