An rock is thrown upward from a platform that is 202 feet above ground at 70 feet per second. Use the projectile formula

h

=

-

16

t

2

+

v

0

t

+

h

0

to determine when the rock hit the ground.

[Recall that

v

0

is the initial velocity of the object and

h

0

is the inital height of the object.]

in 6.36004793262 seconds

Step-by-step explanation:

h (t) = - 16t^2 + v*t + h

0=-16t^2 + 70*t + 202

t = (-70 ± sqrt ((70^ (2) - 4 * (-16) * 202)) / (2 * (-16))

t = (-70 ± sqrt (4900+64*202)) / -32

t = (-70 ± sqrt (4900+12928)) / -32

t = (-70 ± sqrt (17828)) / -32

t = (-70 ± sqrt (2*2*4457)) / -32

t = (-70 ± 2 sqrt (4457)) / -32

t = (-70 + 2 sqrt (4457)) / -32

t=-1.98504793262

t = (-70 - 2 sqrt (4457)) / -32

t=6.36004793262

since we need the time it will fall to the ground after it was thrown and time can't be negative, answer is

in 6.36004793262 seconds

I don't know till what I am supposed to round, so I will leave it there.