Projectile motion - about baseball

Question

A batter hits a fly ball which leaves the bat 0.90 m above the ground at an angle of 61 degrees with an initial speed of 26 m/s heading toward centerfield. Ignore air resistance.

The ball is caught by the centerfielder who, starting at a distance of 1005 m from home plate, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed

Answer 1

What do you mean " makes the catch at ground level" ?

Do you mean he is picking the ball up off of the ground? Because if the ball is at ground level, then you have not made the catch.

Clarify

Answer 2

The range of a projectile is given by the formula

R = V^2(sin 2A)/g

where

R = range (maximum horizontal distance travelled by projectile)

V = initial velocity = 26 m/sec (given)

A = angle of launch = 61 deg (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting values,

R = 26^2(sin 2*61)/9.8

R = 58.5 meters

NOTE --- this is the maximum horizontal distance that the ball travelled. At this point, it is still 0.9 m above the ground hence it will still take the ball an additional "t" seconds before it hits the ground.

Y = Vy(t) + (1/2)gt^2

where

Vy = vertical component of the velocity = 26(sin 61) = 22.74

and all the other terms have been previously defined.

Substituting appropriate values,

0.9 = 22.74t + (1/2)(9.8)t^2

Rearranging the above equation,

4.9t^2 + 22.74t - 0.9 = 0

and using the quadratic formula,

t = 0.04 sec

Let

X = additional horizontal displacement of the ball

X = Vx(t)

where

Vx = 26(cos 61)(0.04)

Vx = 0.50 m

Total horizontal displacement of ball = 58.5 + 0.50 = 59 meters

Let us now consider the efforts of the centerfielder:

He has to run (1005 - 59) = 946 meters to be able to catch the ball and his total time to negotiate the distance is the total time the ball was on the air (before hitting the ground).

Let

T = total time ball was in the air

T = time ball took to negotiate its range + 0.04

Time ball took to negotiate its range is given by the formula (from your Physics book on Projectiles) = 2V*sin A/g

Substituting appropriate values,

T = time ball took to negotiate its range + 0.04

T = 2(26)(sin 61)/9.8 + 0.04

T = 4.64 + 0.04

T = 4.7 sec.

So, for the centerfielder to catch the ball, his speed should be

V = 946/4.7 = 201.28 m/sec

(which is actually ridiculously incredible, unless he is bionic. Even the great Wiilie Mays or Mickey Mantle, during their primes, could not run this fast!!! Even Barry Bonds on steroids could not do this!!!)

Editorializing aside, this is the procedure on how to analyze/solve this problem.

Anyway, I suggest you check on your given data because first of all, the centerfielder could not be 1005 meters away from the batter and still be inside the park. There is no baseball park that is this big.