In the abscense of air resisitance a ball player tosses a ball straight up how musch speed does the ball increase ascending decrease descending

Question

In the absence of air resistance , a ball player tosses  a ball up straight up, how much does the speed of the ball decrease while ascending, increase while ascending,how does the time of ascent compare with the time of descent.

Answer 1

In the absence of air resistance , a ball player tosses a ball up straight up,

 

how much does the speed of the ball decrease while ascending, increase while ascending,

 

> The ball's speed decreases to zero while ascending; it doesn't increase while ascending.

 

how does the time of ascent compare with the time of descent.

 

> Assuming the ball player catches the ball at the same height as s/he released it, and neglecting air resistance, the times are the same.

Answer 2

Hi Richard;

The ball is launched...

The rate of decrease of speed is...g=-9.80 m/s².

g is the acceleration or deceleration due to gravity.

The rate of speed will decrease 9.80 meters-per-second, every second.

Eventually, the ball will reach its maximum height.  At such point, its velocity is zero.  Then, as the ball descends, it will accelerate as the rate of...g=+9.80 m/s².

 

Answer 3

This is a standard physics question.  The acceleration (g) will be nearly constant.  Technically, g will change as a function of distance from the center of the Earth, but change in 'g' for a ball toss on Earths surface will be extraordinarily negligible.  If you were near a black hole, the gradient of acceleration is HUGE and could not be considered constant.

 

With constant accel 'g' the equations of motion are simple as a function of time (t)

 

• a(t)=-g   Acceleration as a function of time a(t)  (note: sign is important.  We define up as '+')
• v(t)=-gt +v(0)  Velocity as a function of time v(t) with initial velocity v(0)
• s(t)=-1/2gt² +v(0)t +s(0)  position as a function of time s(t) with initial position (let s(0)=0)

 

Currently we have the velocity parametric with time (t) in the second relation. The speed of the ball (wrt time) will linearly decrease to zero (as the ball ascends) and then linearly increase (in neg dir) as the ball descends. It is a linear function because 'g' is constant.  Position with time will not be linear.  It will be parabolic since it changes with the square of time.   When the ball is back at the pitchers hand (s=0), the 'speed' will be identical to the initial upward speed.  The 'velocity' is a vector and therefore will have a negative sign in front, but the same scalar constant speed v(0).

 

You asked how the speed (velocity) of the ball changes during ascent and descent.  It is also possible to solve for velocity as a function of distance by manipulating the 2nd and 3rd equations.