Question

1. A stone is tossed upward with a velocity of 8m/s from the edge of a cliff 63 m high. How long will it take the stone to hit the ground at the foot of the cliff?

Using antidifferentiation method, I got dv=-9.8dt. The I integrated and got v+constant1=-9.8t+constant2. Then simplified to v=-9.8t+constant. Then I plugged 9m/s into v, and 63 into constant to solve for t, which is 5.6s. My answer is wrong, and I don't know how to solve this.

Answer 1

In addition, could I also get help with this substitution question?
(y^3+6y^2+12y+8)(y^2+4y+4)dy

I have no idea. I made the first expression into u, derived it, plugged it back in to get 1/3|integration|(u)du. I don't know how to go on from here.

Answer 2

we know a = -9.8
v = -9.8t + c
when t = 0 , v = 8
8 = 0+c --> c = 8

v = -9/8t + 8

s = -4.9t^2 + 8t + 63 , where s is the distance in metres.

at hitting the ground, s = 0
-4.9t^2 + 8t + 63 = 0
t = (-8 ± √1298.8)/-9.8
= appr 4.49 seconds or a negative t

Answer 3

∫(1/3)u du = (1/3) * 1/2 u^2 = 1/6 u^2 + C

Answer 4

a(t) = -9.8
v(t) = -9.8t + c

Initial Velocity = -8m/s. So,
-8 = -9.8(0) + c
-8 = c

So, v(t) = -9.8t - 8
s(t) = -4.9t^2 - 8t + d

Initial Position = 63m
63 = -4.9(0)^2 - 8(0) + d
63 = d

So, s(t) = -4.9t^2 - 8t +63
0 = -4.9t^2 - 8t +63
t approximately equals = 2.86 or 2.9 seconds
(verified with the textbook answer)