These are always fun, aren't they?A pitched ball is hit by a batter at a 45 degree angle and just clears the outfield fence, 98 m away. Assume that the fence is at the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.

First of all, NEVER take an equation from another problem and just assume it holds. You should know WHY the equation works.

The first thing to do is set an origin and coordinate system. I'll set the origin to be where the ball leaves the bat and let +x be in the direction of the horizontal component of the initial velocity and let +y be upward.

Now set up a table of some kind, indicating what we know:

t0 = 0 s (The time the ball left the bat)

x0 = 0 m

y0 = 0 m

(These indicate the initial position of the ball)

v0x = v0*cos(45)

v0y = v0*sin(45)

(These indicate the components of the initial velocity, v0)

x = 98 m

y =0 m

(These indicate the final position of the ball, at the fence.)

ax = 0 m/s^2

ay = -9.8 m/s^2

(These indicate the components of the acceleration of the ball)

The unknowns are t (the time the ball gets to the fence), vx and vy (the components of the velocity when the ball gets to the fence), and of course v0 (the initial speed of the ball.)

We need an equation with v0 in it. We have eight of them to choose from:

x = x0 + v0x*t + (1/2)ax*t^2

vx = v0x + ax*t

vx^2 = v0x^2 + 2ax*(x - x0)

x = x0 + (1/2)(v0x + vx)t

y = y0 + v0y*t + (1/2)ay*t^2

vy = v0y + ay*t

vy^2 = v0y^2 + 2ay*(y - y0)

y= y0 + (1/2)(v0y + vy)t

The first four equations really reduce to only one, since ax = 0 m/s^2:

x = x0 + v0x*t + (1/2)ax*t^2

with the condition that vx = v0x.

So

98 = v0*t*cos(45)

Now, we've got two unknowns in this equation, so we need another equation with only these two unknowns. The first y equation suits the bill:

y = y0 + v0y*t + (1/2)ay*t^2

or

0 = v0*t*sin(45) - 4.9t^2

0 = v0*sin(45) - 4.9t

So our two equations to solve simultaneously are:

98 = v0*t*cos(45)

0 = v0*sin(45) - 4.9t

Solve the bottom equation for t:

t = [sin(45)/4.9]*v0

and insert this into the top equation:

98 = v0*{[sin(45)/4.9]*v0}*cos(45)

98 = [sin(45)*cos(45)]*[1/4.9]*v0^2

So

v0 = sqrt{(98*4.9)/(sin(45)*cos(45)} = 30.99 m/s

(We choose only the + value for the square root since v0 in these equations is a speed, and thus has a positive value.)

-Dan