When Babe Ruth hit a homer over the 12-m-high right-field fence 100m
from home plate, roughly what was the minimum speed of the ball when
it left the bat? Assume the ball was hit 1.0 m above the ground and
its path initially made a 40 degree angle with the ground.

(Ans 34 m/s)

First: list what is given with suitable notation and units.Return to physics page

hf=12m {height of fence}

sf=100m {distance to fence}

theta=40deg {angle with horizon}

g=9.8m/s^2 {acceleration due to gravity}

Second: list the unknown with suitable units.

v = ? m/s {speed of ball}

Third: draw a diagram and decide on a basic equation.

Here think of relationships and what we know; try to have the unknown in the basic equation. Since we know a lot about the horizontal distance, I will choose to begin with

cos(theta)=vh/v {where vh is the horozontal component of speed v}

since vh=sh/t {where sh is horizontal distance and t is time}

we could find v if we could find sh and t.

We can approximate sh for now by seeing that it is 100m plus the distance the ball is beyond the fence when it is one meter above ground; this can be approximated for now by seeing that the distance beyond the fence is a little greater than 11m/tan(theta) from the definition of the tangent function. Thus sh=100m + 11m/(0.8391)=113.1m

To find t we realize that the ball is going both vertically and horizontally and that these components can be treated independently, so since vf = vi + g*t where vf=0 and vi=vv=v*sin(theta) {where vv is the vertical component of the speed of the ball} we have

v*sin(theta)=g*t

and so t=v*sin(theta)/g

Now this is only the time to go up to maximum height; the ball also has to come down to 1.0 meter above ground so t = 2*v*sin(theta)

but vh=sh/t where vh=v*cos(theta)

and so t=sh/v*cos(theta)

and setting both equations equal to t equal to one another, we have

sh/(v*cos(theta))=2*v*sin(theta)/g

multiply both sides by v to get

sh/cos(theta)=2*v*v*sin(theta)/g

multiply both sides by g to get

g*sh/cos(theta)=2*v*v*sin(theta)

multiply both sides by 1/(2*sin(theta)) to get

g*sh/(2*(cos(theta))*(sin(theta)))=v*v

take the square root of both sides and use symmetric property to get

v = sqrt(g*sh/(2*(cos(theta))*(sin(theta))))

This is step four, the working equation!

Fifth: Substitute the values!

v = sqrt(9.8 m/s^2 * 113.1 m / (2 * 0.76604 * 0.64279))

v = sqrt(1108.38 m^2/s^2 / 0.98481)

v = 33.548 m/s

This rounds to v = 34 m/sA better way to find the distance the ball is beyond the fence when it is 1.0 meter above the ground is left to the student and his or her mathematics teacher!

Note: This problem is from Giancoli's 3rd chapter which is on
kinematics and relies on the six formulae:

v = s/t

a = (vf - vi)/t ==> vf = vi + a*t

v = (vi + vf)/2

s = vi*t + a*t^2/2

vf^2 = vi^2 + 2*a*s

where v is average speed, s is distance, a is acceleration, t is time, vi is initial speed, vf is final speed.