Question

I need this ASAPL

i need help ASAP:
a projectile is thrown upward so that its distance above the ground after t seconds is h=-15t^2+510t. After how many seconds does it reach its maximum height?

Answer 1

@campbell_st

Answer 2

derivatives

Answer 3

you familiar with derivative ?

Answer 4

nope

Answer 5

find the vertex

Answer 6

ok here is algebraic way.

Answer 7

just find the line of symmetry which will be t = -b/2a

Answer 8

line of symmetry = 17

Answer 9

b = 520 and a = -15

subsititute and and evaluate

Answer 10

h=17?

Answer 11

t=17 not h.

Answer 12

put this value in
h=-15t^2+510t.
to get h

Answer 13

no i meant h from h=-15^2+510

Answer 14

oh sorry

Answer 15

No need to find h. They just want the value of t at it's peak.

Answer 16

you have found the time after launch that the projectile reaches its maximum height... The max height is found by substituting t = 17 into your equation.

Answer 17

so 17 seconds?

Answer 18

yes

Answer 19

yes use t=17 in the
h=-15t^2+510t.

Answer 20

thanks soo much!! :D

Answer 21

i dont ned that..i just need t

Answer 22

need*

Answer 23

You can find the height if you want, but the problem is only asking for the time at the max height (whatever it is, we don't need to find it)

Answer 24

yea i know..thanks!! :)