Calculating the Center Fielder's Average Speed

[Cheddar]

Homework Statement

A baseball is hit into the air at an initial speed of 36.6m/s and an angle of 50degrees above the horizontal. At the same time, the center fielder starts running away from the batter and catches the ball 0.914m above the level at which it was hit. If the center fielder is initially 110m from home plate, what is his average speed?

Homework Equations

final velocity = initial velocity + (acceleration * time)
displacement = 1/2 (initial velocity + final velocity) time

The Attempt at a Solution

It seems to me that there is some information missing:

It says the fielder catches the ball 0.914m above the level at which it was hit, but it doesn't say how high the ball was when it was hit.

It also says the fielder is initially 110m away from home plate, but doesn't say how far away he is when he makes the catch or how far he runs backwards from the initial position.

 

[tms]

It seems to me that there is some information missing:

It says the fielder catches the ball 0.914m above the level at which it was hit, but it doesn't say how high the ball was when it was hit.

You can put the origin of your coordinate system anywhere. It simplifies things to put it at the point the bat hits the ball.

It also says the fielder is initially 110m away from home plate, but doesn't say how far away he is when he makes the catch or how far he runs backwards from the initial position.

That's one of the things you have to figure out to get the answer.

 

[Cheddar]

Can't figure out how...

 

[rl.bhat]

You can use the formula for projectile motion.
y = x*tanθ - 1/2*g*x^2/2*v^2*cos^2θ. Find x. From that find t.
110 - x is the distance moved by the fielder in time t.

 

[Cheddar]

Okay, so now I have:
initial velocity (vertical) = 28.04 m/s
time (to max height) = 2.86 sec
y (max height) = 120.32 m
horizontal displacement = 134.6 m

So now I'm stuck on the part about the fielder catching the ball 0.914m above the level at which the ball was hit. I believe the horizontal displacement (134.6 m) is the point at which the ball will hit the ground, right? So how do I find where it will be when it is 0.914m above the initial contact point?

 

[tms]

You have an equation that gives y as a function of the initial conditions and t. Solve the equation for t, and plug in the initial conditions and the final value of y. You should find two solutions, one for the ball going up and the other for the ball going down. Then use the t you just found to find the x position of where the ball was caught. Then use t and x to find the average speed of the fielder.