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Q: Determine the speed at which a baseball must move in order for air resistance to be significant?

A: mg = ½ * Cd * A * pair * v^2 pair = density of air (1.2kg/m^3)

m = mass .1488kg

g = 9.8m/s^2

A=(pi*r^2)=3.14*.115m^2=.04154 m^2

Cd = .5 (sphere)

v= square root of ((2*.1488kg*9.8m/s^2)/(.5* 3.14*.04154 m^2* *1.2kg/m^3 ))

v = 36 m/s or 80mph

Q: If the ball were launched vertically into the air at this speed, how high will it go?

A: After finding the speed in the previous question; I would use this equation to find height:

KE = PE

PE = mgy

KE = 1/2mv^2

(½)(.1488kg)(36m/s)^2 = (.1488kg)(9.8m/s^2)y

66.122m=y

Q: how long will it be air born?

x = x0 + v0 *t + 1/2at^2

66.122m = 0m + 36m/s * t + -9.8m/s^2t

v =

I am stuck on this part and I don’t know if the rest of it is correct??? Please help!