Question

A falling stone is at a certain instant 100 feet above the ground. Two seconds later it is only 20 feet above the ground. If it was thrown down with an initial speed of 7 feet per second, from what height was it thrown?

Answer 1

This is a calculus problem not phsyics

Answer 2

. I am assuming that you drop the stone from rest, not throw it down or up.

ok falling objects accelerate to the ground at the constant rate 32 ft/s^2

drop object from some height at time 0. After t secs, since the object is dropped from rest, the speed of the object is v= 32t ft/s

distance object has travelled at time t from the spot you dropped it is
s=16t^2 ft
(how did i get this? distance is the area under the speed graph. so i integrated 32t)

a.
say the object is dropped from height h.
then when it is 100 ft above ground, it has travelled h-100 ft from dropping point.
how much time passed?
h-100 = 16t^2 ----> t= 0.25*√(h-100)

2 seconds later it is only 16 ft above ground (so it has travelled h-16 ft)
h-16 = 16(t+2)^2
h-16 = 16*(0.25*√(h-100)+2)^2

this is an equation in h..... solve it! if you're allowed to use a graphing calculator then its easy.

Answer 3

The answers are 6479/64

Answer 4

6415/64

Answer 5

6607/64

Answer 6

let H(t) = yo -7(t)-16t^{2}

H(c) = 100 = yo -7(c) -16c^{2}
H(c+2) = 20 = yo -7(c+2) - 16(c+2)^{2}


eq1)
20 = yo -7c -14 - 16(c^{2}+4c + 4)
yo = 20 +14 +7c + 16c^{2} +64c +64

eq2)

yo = 100 + 7c +16c^{2}


solve
eq1=eq2
100 + 7c + 16^{2} = 34 + 7c + 16c^{2} +64c+64

2 = 64c

c = 1/32

now

H(1/32) = 100=yo - 7(1/32) - 16(1/34)^{2}

Answer 7

yo = 100 15/64
= 6415/64

Answer 8

that part up there is -16(1/32)^{2

Answer 9

Thankh