A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 22.0 m/s. The cliff is h = 38.0 m above a flat horizontal beach, as shown in the figure below.
How long after being released does the stone strike the beach below the cliff?
s
With what speed and angle of impact does the stone land?
m/s
° below the horizontal
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The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45°. With what speed must the animal leave the ground to reach that height?
Correct: Your answer is correct. m/s
See previous post: Thu,2-4-16, 11:20 AM.
To determine how long it takes for the stone to strike the beach, we can use the equation of motion for vertical displacement. Since the stone is initially thrown horizontally, there is no initial vertical velocity. The equation for vertical displacement is:
h = ut + (1/2)gt^2
Where:
h = vertical displacement (38.0 m in this case)
u = initial vertical velocity (0 m/s in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Rearranging the equation, we get:
t = √(2h/g)
Substituting the given values into the equation, we can calculate the time it takes for the stone to strike the beach:
t = √(2 * 38.0 / 9.8) = √(76.0 / 9.8) = √7.755 = 2.78 s (rounded to two decimal places)
Therefore, the stone takes approximately 2.78 seconds to strike the beach below the cliff.
To determine the speed and angle of impact, we can use the horizontal and vertical components of the stone's velocity. Since the stone is thrown horizontally, the initial horizontal velocity remains constant throughout its motion. The initial horizontal velocity is:
vx = 22.0 m/s (given)
Since there is no initial vertical velocity, the stone only experiences vertical motion due to the gravitational acceleration. The final vertical velocity can be found using the equation:
v = u + gt
Where:
v = final vertical velocity (unknown)
u = initial vertical velocity (0 m/s in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (2.78 s in this case)
Rearranging the equation, we get:
v = gt
Substituting the values, we can calculate the final vertical velocity:
v = 9.8 * 2.78 = 27.244 m/s (rounded to three decimal places)
Now we can determine the speed and angle of impact. The speed is the resultant velocity, which can be found using the Pythagorean theorem:
speed = √(vx^2 + v^2) = √(22.0^2 + 27.244^2) = √484 + 743.034 = √1227.034 = 35 m/s (rounded to two decimal places)
The angle of impact below the horizontal can be found using trigonometry. We have the horizontal and vertical components of the velocity, and we need to find the angle θ such that:
tan(θ) = vertical component / horizontal component
θ = atan(vertical component / horizontal component) = atan(v / vx) = atan(27.244 / 22.0) = atan(1.238) = 51.60° (rounded to two decimal places)
Therefore, the stone lands with a speed of approximately 35 m/s and at an angle of 51.60° below the horizontal.