Question

The height, s, of a ball thrown straight down with initial speed 64 ft/sec from a cliff 80 feet high is s(t) = –16t2 – 64t + 80, where t is the time elapsed that the ball is in the air. What is the instantaneous velocity of the ball when it hits the ground?
256 ft/sec
–96 ft/sec
0 ft/sec
112 ft/sec

Answer 1

@wio

Answer 2

Figure out the time it hits the ground, then plug that into the derivative.

Answer 3

It hits the ground when \(s(t)=0\).

Answer 4

how do i do that

Answer 5

@wio

Answer 6

@arnavguddu

Answer 7

@Lyrae

Answer 8

@eliassaab

Answer 9

@wio

Answer 10

@PFEH.1999

Answer 11

have u studied derivative?

Answer 12

@familyguymath

Answer 13

yas

Answer 14

@PFEH.1999

Answer 15

well then,
\[v(t) = \frac{ d }{ dx } s(t) = -32t - 64 \]
\[a(t) = \frac{ d }{ dx } v(t) = -32 \]
and use this formula:
\[v^2 - v_{0}^2 = 2a \Delta x\]