A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 15.0 m/s. The cliff is h = 66.0 m above a flat horizontal beach.
How long after being released does the stone strike the beach below the cliff?
I got 3.67 s.
With what speed and angle of impact does the stone land?
I got 38.97 m/s for the speed, but I cannot figure out the angle. It says that the answer is in degrees "below the horizontal."
I calculated x to be 55.23m and I tried to calculate -y, but I keep getting the wrong angle. After getting -y I though I was supposed to do arctan(opp/adj) to get the angle "below the horizontal."
Please help! Thanks!
T = 3.67 s is correct for the time to impact. When thowing horizontally, the speed does not affect the time to hot the ground.
Using conservation of energy for the speed Vf at impact,
(1/2) Vf^2 = Vo^2/2 + g H
(1/2) Vf^2 = (15^2)/2 + 9.8*66
= 759.3 m^2/s^2
Vf = 38.97 m/s
The tangent of the angle at impact, measured below the horizontal, is Vy/Vx at impact
theta = arctan (g*T)/15 m/s
= arctan 35.97/15 = 67.4 degrees
Thanks!
It is also arcsin g*T/38.97, where 38.97 is the speed. This gives you the same angle.
To find the time it takes for the stone to strike the beach, you can use the formula for horizontal distance:
\[d = v_x \times t\]
where \(d\) is the horizontal distance traveled by the stone, \(v_x\) is the horizontal component of the stone's velocity, and \(t\) is the time it takes for the stone to strike the beach. In this case, since the stone is thrown horizontally, the initial vertical velocity is 0 and there is no vertical acceleration, so the horizontal component of the velocity remains constant throughout the motion. Therefore, we can set \(v_x\) as the initial velocity of the stone, which is 15.0 m/s.
Now, we know that \(d = 55.23\) m, as you calculated. Rearranging the formula, we can solve for \(t\):
\[t = \frac{d}{v_x}\]
Substituting the given values, we have:
\[t = \frac{55.23\, \text{m}}{15.0\, \text{m/s}} = 3.682\, \text{s}\]
Therefore, the time it takes for the stone to strike the beach below the cliff is approximately 3.682 s.
To find the speed and angle of impact, we can split the stone's motion into horizontal and vertical components.
The stone's vertical motion is governed by the equations of motion under constant acceleration. The only force acting in the vertical direction is gravity. Since the stone starts with an initial vertical velocity of 0 m/s and falls a distance of 66.0 m, we can use the following equation to find the time of flight:
\[h = \frac{1}{2} g t^2\]
where \(h\) is the height of the cliff (66.0 m), \(g\) is the acceleration due to gravity (approximately 9.8 m/s²), and \(t\) is the time of flight.
Rearranging the equation, we get:
\[t^2 = \frac{2h}{g}\]
Substituting the given values, we have:
\[t^2 = \frac{2 \times 66.0\, \text{m}}{9.8\, \text{m/s}^2} = 13.47\, \text{s}^2\]
Taking the square root of both sides, we find:
\[t = \sqrt{13.47\, \text{s}^2} = 3.672\, \text{s}\]
This confirms the time we found earlier.
To find the vertical component of the velocity at impact, we can use the equation:
\[v_y = g \times t\]
Substituting the given values, we have:
\[v_y = 9.8\, \text{m/s}^2 \times 3.672\, \text{s} = 36.004\, \text{m/s} \approx 36.0\, \text{m/s}\]
Now, we can use the horizontal component of the velocity (15.0 m/s) and the vertical component of the velocity (36.0 m/s) to find the speed and angle of impact using vector addition.
The speed can be calculated using the Pythagorean theorem:
\[v = \sqrt{v_x^2 + v_y^2}\]
Substituting the given values, we have:
\[v = \sqrt{(15.0\, \text{m/s})^2 + (36.0\, \text{m/s})^2} = 38.904\, \text{m/s} \approx 38.9\, \text{m/s}\]
Therefore, the speed of impact is approximately 38.9 m/s.
To find the angle of impact, we can use trigonometry. The angle we need is the angle "below the horizontal." Since we have the vertical and horizontal components of the velocity, we can use the inverse tangent function to find the angle:
\[\theta = \arctan\left(\frac{v_y}{v_x}\right)\]
Substituting the given values, we have:
\[\theta = \arctan\left(\frac{36.0\, \text{m/s}}{15.0\, \text{m/s}}\right) \approx 66.62^\circ\]
Note that the angle is above the horizontal, but we want the angle below the horizontal. Therefore, we can subtract this angle from 180° to get:
\[\text{angle below horizontal} = 180° - 66.62° = 113.38°\]
Therefore, the stone lands with a speed of approximately 38.9 m/s and an angle of impact of approximately 113.38° below the horizontal.