Use the model for projectile motion, assuming there is no air resistance and g = 32 feet per second per second. a baseball, hit 3 feet above the ground, leaves the bat at an angle of 45° and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. what is the initial speed of the ball, and how high does it rise? (round your answers to two decimal places.) initial speed ft/s maximum height ft

The initial speed of the ball is 98 ft/s and the maximum height is 75 ft.

Explanation:

Hi there!

The position and velocity vectors of the ball at a time "t" can be calculated as follows:

r = (x0 + v0 · t · cos θ, y0 + v0 · t · sin θ + 1/2 · g · t²)

v = (v0 · cos θ, v0 · sin θ + g · t)

Where:

r = position vector of the ball at time t.

x0 = initial horizontal position.

v0 = initial speed of the ball.

t = time.

θ = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity.

v = velocity vector at time t.

Let´s place the origin of the system of reference at the point at which the ball leaves the bat so that x0 and y0 = 0.

The final position of the ball is the vector

rf = (300 ft, 0) (see attached figure).

Then:

300 ft = x0 + v0 · t · cos θ

0 ft = y0 + v0 · t · sin θ + 1/2 · g · t²

Replacing with the available data:

300 ft = v0 · t · cos 45°               (x0 = 0)

0 ft = v0 · t · sin 45° -1/2 · 32 ft/s² · t²

Solving for v0 · t in the first equation:

300 ft / cos 45° = v0 · t

Replacing v0 · t for (300 ft/ cos 45°) in the second equation and solving it for t:

0 = (300 ft / cos 45°) · sin 45° - 16 ft/s² · t²

0 = 300 ft - 16 ft/s² · t²

-300 ft / -16 ft/s² = t²

t = 4.3 s

With this time, we can calculate v0:

300 ft / (cos 45° · 4.3 s) = v0

v0 = 98 ft/s

The initial speed of the ball is 98 ft/s.

Notice looking at the figure that, at the maximum height, the vertical component of the velocity, vy, is zero (the velocity vector is horizontal). Then using the equation of the vertical component of the velocity vector at the maximum height:

vy = v0 · sin θ + g · t

0 = 98 ft/s · sin 45° - 32 ft/s² · t

-98 ft/s · sin 45° /-32 ft/s² = t

t = 2.2 s

Now let´s calculate the vertical position when t =2.2 s

ry = y0 + v0 · t · sin θ + 1/2 · g · t²

ry = 98 ft/s · 2.2 s · sin 45° - 1/2 · 32 ft/s² · (2.2 s)²

ry = 75 ft

The maximum height of the ball is 75 ft.

The same result is obtained if we use energy conservation:

All the initial vertical kinetic energy (KE) will be converted into potential energy (PE) at the maximum height:

KE = PE

1/2 · m · (v0y)² = m · g · h

1/2 · m · (98ft/s · sin 45°)² = m · 32 ft/s² · h

Solving for h (height) (the mass cancel):

1/2 · (98ft/s · sin 45°)² / 32 ft/s² = h

h = 75 ft

It is quicker this way!