A suspension bridge is 66.0 m above the level base of a gorge. A stone is thrown or dropped from the bridge. At the location of the bridge, g has been measured to be 9.83 m/s2. In each of the following, assume that there is no air resistance.
(a) If you drop the stone, how long does it take for it to fall to the base of the gorge?
s
(b) If you throw the stone straight down with a speed of 22.5 m/s, how long before it hits the ground?
s
(c) If you throw the stone with a velocity of 22.5 m/s at 25.0° above the horizontal, how far from the point directly below the bridge will it hit the level ground?
do u go to wayne state?
a. d = 0.5gt^2,
66 = 0.5 * 9.83 t^2,
66 = 4.915t^2,
t^2 = 66 / 4.915 = 13.4,
t = sqrt(13.4) = 3.66 s.
b. d = Vo*t + 0.5gt^2,
66 = 22.5t + 0.5 * 9.83t^2,
66 = 22.5t + 4.915t^2,
4.915t^2 + 22.5t - 66 = 0,
Use Quadratic Formula to find t:
t = (- 22.5+- sqrt(506.25 + 1297.56))/
9.83,
t = (-22.5 +- 42.47) / 9.83,
t = 2.03 or -6.61, select positive t:
t = 2.03 s.
c. y = Ver.=22.5 m/s * sin25 = 9.5m/s
d(up) = d(down),
Vo*t = 0.5gt^2,
9.5t = 0.5 * 9.83t^2,
9.5t = 4.915t^2
4.915t^2 - 9.5t = 0,
Factor out t:
t(4.915t - 9.5) = 0,
t = 0: 4.915t - 9.5 = 0, 4.915t = 9.5,
t = 1.93 s (Ver),
d = Ver. = 9.5 m/s * 1.93 s = 18.4 m,
d = R = 18.4 m /sin25 = 43.5 m,
d = Hor. = 43.5 m * cos25 = 39.5 m = distance stone lands.
To solve each part of the problem, we can use the equations of motion under constant acceleration. In this case, the stone is under the influence of gravity, so the acceleration is equal to the acceleration due to gravity, g.
Let's solve each part step-by-step:
(a) If you drop the stone, how long does it take for it to fall to the base of the gorge?
When you drop the stone, it falls freely under the influence of gravity. The stone is not initially given any initial velocity. The equation we can use to solve for time is:
s = (1/2)gt^2
Where:
s = distance traveled (66.0 m in this case)
g = acceleration due to gravity (9.83 m/s^2)
Rearranging the equation to solve for time (t), we get:
t = sqrt(2s/g)
Now, we can substitute the values into the equation:
t = sqrt(2 * 66.0 / 9.83)
t = sqrt(132 / 9.83)
t = sqrt(13.4307)
t ≈ 3.66 s
Therefore, it takes approximately 3.66 seconds for the stone to fall to the base of the gorge.
(b) If you throw the stone straight down with a speed of 22.5 m/s, how long before it hits the ground?
In this case, the stone is thrown straight down with an initial velocity of 22.5 m/s, and it falls under the influence of gravity. The equation we can use to solve for time is:
s = ut + (1/2)gt^2
Where:
s = distance traveled (unknown in this case)
u = initial velocity (22.5 m/s)
g = acceleration due to gravity (9.83 m/s^2)
Since the stone is being thrown downwards, the initial velocity (u) is negative (-22.5 m/s).
The equation can be rearranged as a quadratic equation in terms of time (t):
(1/2)gt^2 + ut - s = 0
We can now substitute the values into the equation:
(1/2) * 9.83 * t^2 - 22.5 * t - 66.0 = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b +/- sqrt(b^2 - 4ac)) / (2a)
Where:
a = 1/2 * g
b = u
c = -s
Substituting the values into the formula, we get:
t = (-(-22.5) +/- sqrt((-22.5)^2 - 4 * 1/2 * 9.83 * -66.0)) / (2 * 1/2 * 9.83)
t = (22.5 +/- sqrt(506.25 + 1308.36)) / 9.83
t = (22.5 +/- sqrt(1814.61)) / 9.83
Now, calculate the value inside the square root:
sqrt(1814.61) ≈ 42.60
t = (22.5 +/- 42.60) / 9.83
Now, we can calculate the two possible values of time:
t1 = (22.5 + 42.60) / 9.83 ≈ 6.72 s
t2 = (22.5 - 42.60) / 9.83 ≈ -2.07 s
Since time cannot be negative in this context, the negative value is not meaningful.
Therefore, it takes approximately 6.72 seconds for the stone to hit the ground.
(c) If you throw the stone with a velocity of 22.5 m/s at 25.0° above the horizontal, how far from the point directly below the bridge will it hit the level ground?
In this case, the stone is thrown with an initial velocity of 22.5 m/s at an angle of 25.0° above the horizontal. We are interested in finding the horizontal distance (range) from the point directly below the bridge to where it hits the level ground.
The horizontal and vertical components of the initial velocity are:
Initial horizontal velocity (Vx) = 22.5 m/s * cos(25°)
Initial vertical velocity (Vy) = 22.5 m/s * sin(25°)
Since there is no initial vertical velocity, the stone will experience free fall (as in part (a)). We can use the equation:
s = ut
Where:
s = horizontal distance traveled (unknown in this case)
u = initial horizontal velocity (Vx)
We can rearrange the equation to solve for distance (s):
s = ut / Vx
Now, we can substitute the values into the equation:
s = (22.5 m/s * cos(25°)) * t / (22.5 m/s * cos(25°))
s = t
Since the vertical motion doesn't affect the horizontal distance, the time taken for the stone to hit the ground (from part (a)) is the horizontal distance.
Therefore, the stone will hit the level ground approximately 3.66 seconds after being thrown.