Question

a building is 10.0m tall.From the top of the building a stone is dropped from rest.when the stone is halfway down a second stone is thrown straight downward from the top of the building.the two stones strike the ground simultaneously. find the initial speed of the second stone

Answer 1

Isn't this rectilinear motion? Physics?

Answer 2

kinematics

Answer 3

good meters
g= -9.8m/s
x_o=10

first solve for t when the first stone reaches 5 m

Answer 4

\[h=-\frac{g t^2}{2}-vt\]

Answer 5

\[+h_0\]

Answer 6

what i did is i found the time for first stone and divided by 2

Answer 7

so i will get T for second stone

Answer 8

cant, acceleration

Answer 9

first determine the time when the first rock reaches half the height of the building
so
5=(1/2)(-9.8)t^2+10
then determine when the first stone reaches the ground
0=(1/2)(-9.8)t^2+10

take the difference of the 2 times

thats how long it takes the 2nd stone to reach the ground

Answer 10

so you then solve for v_o of the 2nd stone

0=(1/2)(-9.8)t^2+v_o*t+10
where t is the difference in time of the first stone

Answer 11

but a=9.8 right not (-)

Answer 12

i use g=-9.8 because its moving in a downwards motion

Answer 13

it doesnt matter if its positive or negative, as long as the directions are the same

it would be -9.8 because we start at a height,
as time passes, the height becomes smaller, thus acceleration must be negative, to reduce the height

Answer 14

it would be positive if you used -10 instead of +10

if you used a postive accelereration, then the initial velocity would have to be positive because they are both travelling in the same direction

Answer 15

i got t=1.01s for the first stone