Answers 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.)

The initial speed of the ball is  98.22 ft/s

The maximum height of the ball is 75.02 ft.

Step-by-step explanation:

Hi there!

The position and velocity vectors of the ball can be calcualted as follows:

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

v = (v0 · cos α , v0 · sin α + g · t)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity vector at time t.

Let´s call r final to the position vector when the ball is caught (see attached figure). The components of r final are:

x = 300 feet

y = 0

Then, r final = (300 feet, 0)

Notice that the origin of the frame of reference is located at the launching point.

Then, when the ball is caught:

300 feet = x0 + v0 · t · cos α

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

We have a system of two equations with two unknowns (t and v0), so, let´s solve it.

Let´s solve the first eqation for v0:

300 feet = x0 + v0 · t · cos α

Since the frame of reference is located at the launching point (x0 = 0). Then:

300 feet = v0 · t · cos 45°

300 feet /( t · cos 45°) = v0

Now, let´s replace v0 in the secod equation of the system:

0 = y0 + v0 · t · sin α + 1/2 · g · t²    (y0 = 0)

0 = 300 feet/(t · cos 45°) · t · sin 45° - 1/2 · 9.8 m/s² · t²

0 = 300 feet · sin 45°/cos45° - 4.9 m/s² · t²       (sin45°/cos45° = 1)

0 = 300 feet - 4.9 m/s² · t²

-300 feet = -4.9 m/s² · t²

-300 feet / -16.08 ft/s²  = t²

t = 4.32 s

Now, we can calculate the initial velocity:

300 feet /( t · cos 45°) = v0

300 feet /( 4.32 s · cos 45°) = v0

v0 = 98.22 ft/s

The initial speed of the ball is  98.22 ft/s

To obtain the maximum height of the ball, we have to find at which time the ball is at its maximum height. We can do this in several ways, but we will use the equation of velocity. See in the figure that when the ball is at its maximum height, the y-component of the velocity is zero (because the vector is horizontal). Then, using the equation of the y-component of the velocity:

vy = v0 · sin α + g · t

0 = 98.22 · sin 45° - 32.15 ft/s² · t

(-98.22 · sin 45°) / - 32.15 ft/s² = t

t = 2.16 s

(Notice that this time is half the total time of flight calculated above (4.32 s). This is also the time at which the horizontal distance traveled by the ball is half the total horizontal traveled distance. This is so because the trajectory of the ball is a parabola).

Now that we have the time, we can calculate the y-component of the position vector (r1y in the figure).

r1y = v0 · t · sin α + 1/2 · g · t²

r1y = 98.22 ft/s · 2.16 s · sin 45° - 1/2 · 32.15 ft/s² · (2.16 s)²

r1y = 75.02 ft

The maximum height of the ball is 75.02 ft.