_Part A_
Let the first stone travel distance x(t) in and velocity u(t) in time t.
Let the second stone travel distance y(t) in and velocity v(t) in time t.
Let the second stone be released at time ŧ after the first.
Given:
x(0) = y(0) = 0[m]
u(0)=0[m/s]
v(ŧ)=26.0[m/s]
ŧ = 1.60[s]
g=9.8[m/s²]
(NB: Measuring distance down from roof.)
Find:
t such that x(t)=y(t)
Using:
x(t) = u(0)t + ½gt²
y(t) = v(ŧ)(t-ŧ) + ½g(t-ŧ)²
So using the given values and x(t)=y(t):
4.9 t² = 26.0×(t-1.60) + 4.9(t-1.60)²
Solve for t.
_Part B_
Use:
x(t) = ½gt²
_Part C_
Use:
u(t) = gt
v(t) = v(ŧ) + g(t-ŧ)
A stone is dropped from the roof of a building; 1.60s after that, a second stone is thrown straight down with an initial speed of 26.0m/s , and the two stones land at the same time.
part a. How long did it take the first stone to reach the ground?
part b.How high is the building?
part c.What are the speeds of the two stones just before they hit the ground?
1 answer
1 answer