Question

A catapult launches a boulder with an upward velocity of 112 ft/s. The height of the boulder, h, in feet after t seconds is given by the function .
How long does it take the boulder to reach its maximum height? What is the boulder’s maximum height? Round to the nearest hundredth, if necessary.

Answer 1

is given by the 'what' function?

Answer 2

oh sorry its y=-16t^2+112t+30

Answer 3

y = height?

Answer 4

yes

Answer 5

differentiate to get expression for velocity ate time t
at maximum height v = 0
so
v = -32t + 112 = 0
solve for t

Answer 6

now plug in this value of t into your formula to get maximum height

Answer 7

do you follow that ok?

Answer 8

32t = 112
t = 3.5 secs
= time to get maximum height

maximum ht = -16(3.5)^2 + 112(3.5) + 30 ft

Answer 9

yes - same result as i got
though they quoted the formula v = u + at

Answer 10

u for initial velocity?

Answer 11

i derived v = u + at from your formula for the height at time t

Answer 12

u = 112 ft/s

Answer 13

yes u - initial velocity

Answer 14

yeah guess they are pretty much the same, oh well :)

Answer 15

there's i believe 4 equations of motion for particle moving at constant velocity

s = (u +v)t / 2
v = u + at
s = ut + (1/2)at^2
v^2 = u^2 + 2as

and are useful for problems of motion under gravity

Answer 16

u is interchangeable with Vo where o is a subscript 0
what is s?

Answer 17

s is distance or more accurately displacement

Answer 18

ahh okay, I know that as (triangle/delta) x or y depending on which vector, same thing though

Answer 19

e.g. if a projectile is dispatched from top of a building 100 ft high displacement is -100
- yea its a vector