but I think I see the problem
you mean
H = -5 t^2 + 20 t + 60
when t = 0, the height is 60. That is the height of the cliff
if t = 2
h = -5(4) + 20(2) + 60
= -20 +40 + 60 = 80
when dH/dt = 0 or vertical speed is 0 it stops going up
0 = -10 t + 20
t = 2 seconds to top
(that is answer d)
at t = 2
we did it already, 80 meters (part b = part c)
A stone was thrown from the top of a cliff. The height, H meters, of the stone above sea level t seconds after it was released is given by
H=-5t+20t+60.
a)How high is the cliff? What makes you say this?
b) How high is the stone after two seconds?
c) What is the maximum height?
d) After how many seconds does it reach this maximum height?
2 answers
what your teacher did was say g =-10 m/s^2
F = m g = m a
so a = -10 m^s^2
then
v = Vi - g t = Vi - 10 t
h = Hi + Vi t - (10/2)t^2
Vi = 20
v = 0 at top = 20 - 10 t
so t at top = 2
h at top = 60 + 40 - 20 = 80
F = m g = m a
so a = -10 m^s^2
then
v = Vi - g t = Vi - 10 t
h = Hi + Vi t - (10/2)t^2
Vi = 20
v = 0 at top = 20 - 10 t
so t at top = 2
h at top = 60 + 40 - 20 = 80
0 answers