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.
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.
thanks, but C is incorrect any idea why?
nevermind, it's suppose to be negative 93.9. thanks
To find the solutions to these questions, we need to analyze the motion of the stone.
(a) To find the maximum height above the window reached by the stone, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (0 m/s when the stone reaches its highest point)
vi = initial velocity (24.0 m/s)
a = acceleration (due to gravity, -9.8 m/s^2)
d = displacement (maximum height above the window to be found)
First, we need to find the time it takes for the stone to reach its highest point. We can use the equation:
vf = vi + at
Substituting the given values:
0 m/s = 24.0 m/s + (-9.8 m/s^2)t
Simplifying the equation:
-24.0 m/s = -9.8 m/s^2t
Dividing both sides by -9.8 m/s^2:
t ≈ 2.45 s
Now, we can find the maximum height using the kinematic equation:
vf^2 = vi^2 + 2ad
Substituting the given values and the time calculated above:
0 m/s = (24.0 m/s)^2 + 2(-9.8 m/s^2)d
Simplifying the equation:
0 m/s = 576.0 m^2/s^2 - 19.6 m/s^2d
Rearranging the equation:
19.6 m/s^2d = 576.0 m^2/s^2
Solving for d:
d ≈ 29.4 m
Therefore, the maximum height above the window to which the stone rises is approximately 29.4 meters.
(b) To find the time the rock is in the air before it reaches the ground, we can use the equation:
d = vi * t + (1/2) * a * t^2
Where:
d = displacement (420 m)
vi = initial velocity (24.0 m/s)
a = acceleration (due to gravity, -9.8 m/s^2)
t = time in the air (to be found)
Substituting the given values:
420 m = 24.0 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation:
420 m = 24.0 m/s * t - 4.9 m/s^2 * t^2
Rearranging the equation:
4.9 m/s^2 * t^2 - 24.0 m/s * t + 420 m = 0
This is a quadratic equation, which can be solved using the quadratic formula. The equation takes the form of ax^2 + bx + c = 0, where:
a = 4.9 m/s^2
b = -24.0 m/s
c = 420 m
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values into the formula:
t = (-(-24.0 m/s) ± √((24.0 m/s)^2 - 4 * 4.9 m/s^2 * 420 m)) / (2 * 4.9 m/s^2)
Simplifying the equation:
t ≈ 17.50s or t ≈ 4.80s
Since time cannot be negative, the stone is in the air for approximately 4.80 seconds before it reaches the ground.
(c) To find the velocity of the stone just before it hits the ground, we can use the equation:
vf = vi + at
Substituting the given values:
vf = 24.0 m/s + (-9.8 m/s^2)(4.80 s)
Simplifying the equation:
vf = 24.0 m/s - 47.04 m/s
vf ≈ -23.04 m/s
Therefore, the velocity of the stone just before it hits the ground is approximately -23.04 m/s.