Question

At a particular baseball stadium the outfield wall is 310 feet away from home plate and the wall is 6 feet tall. A player hit the ball when it was 3 feet off of the ground with an initial upward velocity of 64 feet per second and an initial outward velocity of 77 feet per second.
Write the function h(x) that gives the height of the ball after time (t).

Answer 1

wait I don't even think you need how far the wall is..guys do you think we would need:
d = Vit + (1/2)at^2 ?

Answer 2

a = -9.8 or g.. amistre64 what do you think.. I'm not sure if this is right..

Answer 3

I'd say:
h(t) = 64t + (1/2) (-9.8) * t^2

Answer 4

V(0) = kt + 64

S(t) = (kt^2)/2 + 64t + C; S(0) = 3

S(t) = (kt^2)/2 + 64t + 3

the 2 velocities has me wondering :)

Answer 5

ohh nice amistre..

Answer 6

outward velocity shouldnt affect the height of the ball; but it does tell us how long it takes to travel the distance, assuming no decrease in velocity ... or the decrease is negligible

Answer 7

we have to attain a height of 6 feet; and a vertex of 310 to clear the wall right?

Answer 8

the upward and outward velocities can be combined some how to form an initial acceleration; or at least a constant acceleration right?

Answer 9

64/77 might be akin to a constant acceleration, but I can recall if that would be accurate :)

Answer 10

if so then k = 64/77 ...... but like i said, its fuzzy

Answer 11

Thank you amistre64!

Answer 12

So based off of this, how long will the ball stay in the air?

Answer 13

we can cheat and use the -16 for k to determine how long it stays airborne... maybe

Answer 14

-16t^2 + 64t + 3 = 0 when t = ?

Answer 15

when t = abt 4.04 its back on the ground

Answer 16

So it stays in the air for... 4.04 seconds?

Answer 17

thats what I get if my reasoning is sound. gravity has the same affect on an object whether its just tossed straight up or if it arcs over a distance; gravity tugs it back to earth at the same constant. so the time it stays airbone would be about 4.04 secs.

Answer 18

Hmm. Good explaining, thank you once again amistre64 :)

Answer 19

your welcome; hope it helps out :)