Batter hits a fly ball which leaves the bat 0.90 m above the ground at an angle of 65° ?

Question

with an initial speed of 25 m/s heading toward centerfield. Ignore air resistance. How far from home plate would the ball land if not caught?

The ball is caught by the centerfielder who, starting at a distance of 105 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

Most of John F's analysis was right.  However, I disagree slightly.  He didn't account for the initial height of 0.9 m.  The 4.62 secs is the time it will take to fall back down to 0.9 m.  To calculate the additional time to hit the ground use t=d/v  or t=0.9m/(25 sin 65)=0.0397 secs.  So, the time will actually be 4.66 secs.  Using his equation, x=xi+vit+1/2 at^2 gives us x=25 cos 65 * 4.66= 49.2 meters.  

The second part of his calculation was incorrect also.  Take the 105 m - 49.2 m =  55.8 meters.  That is how far he is traveling to meet the ball.  So, his velocity would be 55.8 m / 4.66 secs or 11.97 m/s.  

Answer 2

Ok, lets consider the vertical motion of the ball from its start till it reaches its highest point.

At the highest point, the vertical velocity = 0 and the time taken is half the original time for the ball to hit the ground.

Using the simple equations of motion, v = u +at

here v = 0, u = 25 (sin 65), a = -9.81

hence t = 2.31 s,

but the time to hit the ground would be 4.624 s

Now considering the horizontal motion of the ball, and using another equation of motion s = ut + 0.5 at^2

in its horizontal motion, a the acceleration = 0,

therefore the distance, s = ut, where u = 25 (cos) 65 and t = 4.624

it follows that s = 48.9 m

as for the fielders speed, the time taken is 4.624 seconds and he runs 105 m, since average speed, v = distance/time

v = 105/4.624 = 22.7 m/s