2 short physics questions?

Question

1) If an object is launched at the same velocity 2 times: one at 30 degrees and one at 33, which one woudl go further?

2) the maximum height of a fly ball is 20m. How much time does the fielder have run to the predicted position of the ball for the catch?

Answer 1

1) The 33 degree will go further. 45 degrees is the optimal angle so the closer to that, the further the object will travel.

2) Are you assuming that the outfielder started running at the crack of the bat, or when the ball reaches the maximum height?
Use d=.5at^2

If you're assuming the outfielder moves when the ball is at maximum height, it will would out like this...

20=.5(9.8)t^2

4.8=t^2

so t= 2.02 sec if the outfielder starts running at the maximim height.

If it's at the crack of the bat, then double the answer to get 4.04 sec.

Answer 2

(1) The one at 33 degrees. The maximum can be achieved at 45 degrees.

(2) There are two equations you need to know. First is v^2 -v0^2 = 2ad and v = v0 + a*t
where v is the velocity, v0 is the velocity at the start, "a" is the acceleration, "d" is the distance, and t is the time.

v = 0 when the ball is at its maximum (that is the height component, the shadow of the ball going in the vertical direction.)
So you have to solve v^2 -v0^2 = 2ad first.
If v = 0 at the maximum height, then
0 - v0^2 = 2ad (Important: a is negative because gravity is pointing the opposite direction as the pop fly).

-v0^2 = 2*(-9.8 m/s^2)*20 m
v0^2 = 2*9.8*20 m^2/s^2
v0 = square root of 398 m^2/s^2
v0 = 19.8 m/s (this is the initial velocity in the vertical direction)

Now use the second equation to get the time.
v = v0 + a*t
v at the maximum height is zero in the vertical direction.
v = 0 = 19.8 m/s + (-9.8 m/s^2) * t
19.8 m/s = (9.8 m/s^2) * t
t = 2 s
So it takes the ball about two seconds to reach its maximum height. It should take another 2 seconds for it to hit the ground.
So the total time in the air is 2*t or 4 seconds.

****
I didn't answer fast enough. Two others answered faster but we all came to the same answer, so I think the fastest wins. On an aside, I still remember solving this one 40 years ago. It was my first "hard" problem in physics. I was so thrilled. Spend some time trying to solve these problems. You will feel really good about yourself when you "get it". I never got into physics after college but I am sure glad I took it. Good luck in life, young man.

Answer 3

1. At first blush I thought it would depend on whether the angle is measured from the horizontal or vertical, but what really matters is which angle is closest to 45 degrees
In either case the angle 33 degrees is the angle that will result in the object going farther.

2. Assume that the ball starts it's trajectory at about 1 meter rises to 20 meters and falls into the glove of the fielder on the ground.
d=1/2*g*t^2
solve for t
tup=sqrt(19*2/9.8)=1.969
tdown=sqrt(20*2/9.8)=2.020
Ttotal=tup+tdown=3.989

I assume that the person that gave you the problem was expecting the problem to be up 20 down 20 and g=10
Such a person would be expecting the answer to be 4