Free-Fall Motion

Question

As a promotion for the Houston Astros downtown ballpark, a competition is held to see who can throw a baseball the highest from the front row of the upper deck seats, 83ft above field level. The winner throws the ball with an intitial vertical velocity of 92ft/sec and it lands on the infield grass. 

A) Find the maximum height of the baseball

B) How much time is the ball in the air?

C) Determine its vertical velocity when it hits the ground?

Answer 1

Since this is one dimensional kinematics (vertical) we can say the ball is thrown upwards initially at 92ft/s. During the flight of the ball, at first it flies upward but is decelerated by gravity until its upward speed is zero (when it reaches max height) and then it falls and gains downward speed due to gravity. 

 

Part A:

In order to find max height, you need to know how long it took gravity to decelerate the ball down to zero speed from the initial speed. Therefore you use the equation v(final) = v(initial)+a*t

 

a=g=-32ft/s^2 

v(i)=92ft/s

v(f)=0ft/s

 

so you have 1 unknown and you solve:

 

0=92-32*t

t=2.875s

 

Then, to final max height you plug the time into the equation for distance traveled under constant acceleration: d=d(i)+v(i)*t+0.5*a*t^2

 

d(i)=83ft

v(i)=92ft/s

a=g=-32ft/s^2

t=2.875s

 

d=83+92*(2.875)-0.5*32*(2.875^2)=215.25ft <-- max height

 

An alternative (and faster method) to do this is to use the equation v(f)^2=v(i)^2+2*a*d

 

v(f)=0

v(i)=92ft/s

a=g=-32ft/s^2

 

0=92^2+2*-32*d=8464-64d

64d=8464

d=132.25ft

 

Now you need to remember to add in the initial height (height of the stands) which is 83ft:

 

132.25+83=215.25ft

 

 

Part B: from Part B, you just solved for how much time the ball spent moving up to max height and what that height is. Now you need to calculate time needed for the ball to fall to earth from that max height. 

 

You again use the equation for distance traveled under constant acceleration but this time, t is the unknown:

 

d(f)=0

d(i)=max height = 215.25ft

v(i)=zero <--- start to fall from zero speed at max height point

a=g=-32ft/s^2

 

so the equation is:

 

0=215.25+0*t-0.5*32*t^2=215.25-16t^2

 

215.25=16t^2

t=3.668s

 

Add this time to the time the ball spent traveling up to get total flight time:

 

t(total)=3.668+2.875=6.543s

 

Part C: you calculated the time spent by the ball in the falling portion, you know it started falling from zero speed and you know the max height (or distance the ball ultimately fell) so use the equation for velocity under constant acceleration:

 

v(f)=v(i)+a*t

 

v(i)=0

a=g=-32ft/s^2

t=3.668s

 

v(f)=-32*3.668=-117.4ft/s <-- the negative sign indicates the ball was falling downwards. The speed is 117.4ft/s

Answer 2

s(t) = -gt² + v₀t + h₀

 

s(t) = -16t² + 92t + 83    vertex is at (h, k)

 

h = -b/2a

h = -92/2(-16)

h = -92/-32

h = 6

 

k = s(2)

k = -16(4) + 92(2) + 83

k = -64 + 182 + 83

k = 201