# Integration problem?

• Jun 3rd 2009, 03:49 PM
Rainy2Day
Integration problem?
This problem is a hw problem that I have to do, but I really don't know how to start it. It would be GREATLY appreciated if anyone could simplify it and explain to me how to find the answer in small steps.

A baseball thrown at 100 mph will be hit at about 130 mph.
a) assuming the ball goes straight up and comes into contact with the bat at 5ft. off the ground, how high could the ball reach?
b) If the ball is hit to the outfield, how much time does the outfielder have to get under the ball from the time the ball was first hit?

• Jun 3rd 2009, 03:58 PM
skeeter
Quote:

Originally Posted by Rainy2Day
This problem is a hw problem that I have to do, but I really don't know how to start it. It would be GREATLY appreciated if anyone could simplify it and explain to me how to find the answer in small steps.

A baseball thrown at 100 mph will be hit at about 130 mph.
a) assuming the ball goes straight up and comes into contact with the bat at 5ft. off the ground, how high could the ball reach?
b) If the ball is hit to the outfield, how much time does the outfielder have to get under the ball from the time the ball was first hit?

first, convert 130 mph into ft/sec ... that will be the initial velocity, v(0).

$a = -32 \, ft/sec^2$

integrate ...

$v = v(0) - 32t$

integrate ...

$h = h(0) + v(0) \cdot t - 16t^2$

when the ball reaches its maximum height, v = 0

assuming the outfielder (? for a ball hit straight up ?) catches the ball at the height it was hit, final height will be h = 5 ft ... use the height equation to solve for t.
• Jun 3rd 2009, 04:50 PM
Rainy2Day
I did what you told me and I got 575.01 ft. as the max height. Is that correct?
• Jun 3rd 2009, 05:23 PM
skeeter
Quote:

Originally Posted by Rainy2Day
I did what you told me and I got 575.01 ft. as the max height. Is that correct?

I get 573 ft ... close enough.
• Jun 3rd 2009, 05:35 PM
Rainy2Day
Woah I actually did something right! (Rofl)

May I know what you got for the time for part b? (Happy)
• Jun 3rd 2009, 07:21 PM
skeeter
Quote:

Originally Posted by Rainy2Day
Woah I actually did something right! (Rofl)

May I know what you got for the time for part b? (Happy)

double the time it took to get to the top of its trajectory.
• Jun 3rd 2009, 07:33 PM
Rainy2Day
so 5.98 seconds x 2 = 11.92 seconds?