Question

A stone is thrown vertically upward from outside a window, with an initial velocity of 24.0 m/s. It moves up, and then down to the ground 420 m below the point where it started. Let `up' be positive.

(a) What is the maximum height above the window to which the stone rises?
________m.

(b) How long is the rock in the air before it reaches the ground?
________ s.

(c) What is its velocity just before it hits the ground?
_______ m/s.

Answer 1

a. hmax = (V^2-Vo^2)/2g.
hmax = (0-576)/-19.6 = 29.4 m.

b. Tr = (V-Vo)/g.
Tr = (0-24)/-9.8 = 2.45 s. = Rise time.
Vo*t + 0.5g*t^2 = 420 + 29.4 = 449.4 m.
0 + 4.9t^2 = 449.4
t^2 = 91.7
Tf = 9.58 s. = Fall time.

Tr + Tf = 2.45 + 9.58 = 12 s. = Time in
air.

c. V = Vo + gt = 0 + 9.8*9.58=93.9 m/s.

Answer 2

thanks, but C is incorrect any idea why?

Answer 3

nevermind, it's suppose to be negative 93.9. thanks