Question

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.]

Answer 1

in 6.36004793262 seconds

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.