# A batter hits a fly ball which leaves the bat 0.98 m above the ground at an angle of 60 degree with initial speed 28 m/s?

a) How far from home plate would the ball land if not caught?

(b) The ball is caught by the centerfielder who, starting at a distance of 103 m from home plate, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed.

m/s

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eTo solve this problem, we need to determine the time the baseball is in the air. In the following work, I will be doing this. The first steps are to determine the vertical and horizontal components of the baseball’s initial velocity.

Vertical = 28 * sin 60 = 14 * √3 m/s

Horizontal = 28 * cos 60 = 14 m/s

During the time the baseball is in the air, it has a vertical acceleration of -9.8 m/s^2. Let’s use the following equation to determine the baseball’s vertical velocity the moment it lands.

4 * -9.8 * -0.98

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

a = -9.8 m/s^2, d = final height – initial height = 0.0 – 0.98 = -0.98 meter

vf^2 = 588 – 2 * -9.8 * -0.98

vf^2 = 588 – 192.08

vf = ±√568.792

This is approximately 23.8 m/s. But since the baseball is falling, the vertical velocity is negative. Let’s use the following equation.

vf = vi + a * t, a = -9.8 m/s^1

-√568.792 = 14 * √3 – 9.8 * t

t = [(-√568.792 – 14 * √3)] ÷ 9.8

The time is approximately 4.9 seconds.

a) How far from home plate would the ball land if not caught?

d = 14 * [(-√568.792 – 14 * √3)] ÷ 9.8

This is approximately 69.7 meters. I am still working.

(b) The ball is caught by the centerfielder who, starting at a distance of 103 m from home plate, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed.

To answer this problem, I must assume that he starting running the second the baseball was hit.

Speed d ÷ = 69.7 ÷ *([(-√568.792 – 14 * √3)] ÷ 9.8

This is approximately 14.3 seconds. I hope this is helpful for you.

• You should thank her. It is certainly not deserved. You claim more precision than the question numbers might suggest possible and completely missing the answer on the second part.You give a speed in units of seconds while the cardinal should be the distance traveled over time...33 m / 5 s

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• Use the general equation for position in time with constant acceleration.

s = s + v₀t + ½at²

In the vertical, with up positive and the ground the origin, the equation would be

0 = 0.98 + (28sin60)t + ½(-9.8)t²

0 = 0.98 + 24.25t - 4.905t²

t = (-24.25 ±√(24.25² - 4(-4.905)(0.98))) / (2(-4.905))

t = - 0.040 s ignore

or

t = 4.989... s is the time the ball is in the air.

in that time, assuming home base is origin and towards outfield is positive, the ball travels horizontally

s = 0 + (28sin60)(4.989) + ½(0)(4.989²)

s = 69.8489...

s = 70 m ANSWER a

Assuming the fielder starts running the moment the ball is hit,

v = d/t

v = (103 - 70) / 4.989

v = 6.6445...

v = 6.6 m/s

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• R = (v₀²/2g)(1 + √(1 + (2gy₀/v₀²sin²θ))) sin(2θ)

R = (28²/19.6)(1 + √(1 + (19.6•0.98/28²sin²60))) sin(120)

R = (40)(1 + √(1 + (19.2/784•0.75))) 0.866

R = (40)(1 + √(1.0326531)) 0.866

R = (40)(1 + 1.016195) 0.866

R = 69.8 m

distance he runs is 103– 69.8 = 33.2 m

but we don't know his starting time...

range of object projected at an angle θ

R = (V²/g)sin(2θ)

V is initial velocity in m/s

g is the acceleration of gravity 9.8 m/s²

Time of flight = R/(Vcosθ)

Max height

h = (V²/2g)sin²θ

R/h = cotθ

different heights:

R = (v₀²/2g)(1 + √(1 + (2gy₀/v₀²sin²θ))) sin(2θ)

d is the total horizontal distance travelled by the projectile.

v is the velocity at which the projectile is launched

g is the gravitational acceleration—usually taken to be

9.81 m/s² near the Earth's surface

θ is the angle at which the projectile is launched

y₀ is the initial height of the projectile

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