physic question

Question

A baseball is thrown vertically into the air with a speed of 24.7 m/s. How high does it go? How long does the round trip up and down require? Use physic symbols.

Answer 1

Good Afternoon Gin L.

This question delves into the concepts of kinematics and projectile motion with the ball in free fall. I mean free fall since I am going to simplify the equation since there is only one major force affecting this ball inside of the question namely weight derived from the acceleration of gravity.

The problem...A baseball is thrown vertically into the air with a speed of 24.7 m/s. How high does it go? How long does the round trip up and down require? Use physic symbols. Looks for the distance traveled to the top of the arc in the throw and the time taken to reach the starting point (our reference frame).

This problem can be solved using two kinematic equations.

Velocity final = velocity initial + (acceleration * time) # Vf = Vo + at

position final = position initial + (velocity initial *time) +((1/2)*acceleration* time^2) #Xf = Xo +Vot +(1/2)at^2

1. Lets look for the time taken for the ball to hit 0m/s finding the time taken of travel to reach the apex of the throw.

Vf (at top) = 0m/s

Vo(starting velocity) = 24.7m/s

a(acceleration due to gravity) = 9.8m/s^2

t(time) = unknown (in s or seconds)

0m/s = 24.7m/s + (-9.8m/s^2 * t(s)) # we need to separate the time in order to solve for time

add 9.8m/s^2*t to both sides...

9.8m/s^2 *t = 24.7m/s

divide by 9.8m/s^2 to isolate time

t = (24.7m/s)/(9.8m/s^2) # this is looking pretty messy what if we tried to place the physics signs in their proper place.

For example (m/s)/(m/s^2) will be flipped based on fraction division rules to (m/s) *(s^2/m)

m/m = 1

s^2/s = s

Therefore, the units is definitely in seconds and we are on the right track.

t = (24.7/9.8) *(s)

t = 2.5204 s #not full but approximation remember to use the full value when solving until you find a solution. In this case the approximation is for viewing to the 4th decimal place

lets plug in the second equation to find the distance traveled to the top since we now have the missing component of time to reach the top...

Xf = X (unknown)

Xo = 0m (initial start reference frame)

Vo = 24.7m/s

a = -9.8m/s^2

t = 2.5204s

Xf = Xo +Vo*t+(1/2)a*t^2 # plug in parts to the equation

X = 0m +(24.7m/s)*(2.5204s) +(1/2)(-9.8m/s^2)(2.5204s)^2

X = 0 + 62.254m + (-31.127m) = 31.127m

Total distance from top of ball throw to the reference frame start = 31.127m

The final part of the question regarding the time taken to reach the initial point in this case is relatively simple. If there are no forces of drag or air resistance, the velocity will be the exact same as going up reverted to being negative. Looking from a conceptual standpoint, the time taken to reach 0 from 24.7 will be equal to the time taken to reach -24.7 from 0.

Therefore, the total air time from beginning to end is 2*t(top) = 5.0408s

I hope this helps with your conceptual grasp of kinematics and free fall of objects Gin L.

Answer 2

Using gravitational acceleration and the definition of acceleration, you can find the time it takes to reach maximum height (what is the velocity at this height?). Total time for round trip is twice this value. Up and then back down.

Select a kinematic equation that only has x for it's unknown and solve for height x. I chose v² = v_o² + 2a(x-x₀), but you can use a different one.