A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 10.6 m/s. The cliff is 25 m above a flat, horizontal beach.
A.) How long after being released does the stone strike the beach below the cliff? The acceleration of gravity is 9.8 m/s^2. Answer in units of s
B.) At impact, what is its speed? Answer in units of m/s
C.)At what angle below the horizontal does it land? Answer in units of ◦
A.) How long after being released does the stone strike the beach below the cliff? The acceleration of gravity is 9.8 m/s^2. Answer in units of s
Well, it seems like the stone had quite the adventure in the air before it makes a beach entrance. To calculate the time it takes to strike the beach below, we need to consider the horizontal motion. Since the stone is thrown horizontally, there is no initial vertical velocity. But worry not, gravity will take care of that during its flight. So, using the equation for horizontal motion, we can find the time:
Time = Distance / Velocity
The distance is the horizontal distance the stone travels, which is the same as the horizontal distance from the edge of the cliff to the point where it strikes the beach. So, ignoring any potential resistance from seagulls or beach police, the distance is 25 meters.
Plugging in the values, Time = 25 m / 10.6 m/s = approx. 2.36 s.
B.) At impact, what is its speed? Answer in units of m/s
Well, well, well, look who's speeding for a grand beach landing! To find the speed at impact, we need to consider the vertical motion. The stone falls vertically under the influence of gravity. Therefore, its speed at impact will be the same as the vertical velocity right before it hits the beach.
Using the vertical motion equation, Vf = Vi + gt, where Vf is the final velocity, Vi is the initial velocity (which is 0 m/s in this case), g is the acceleration due to gravity (9.8 m/s^2), and t is the time in the air.
Plugging in the values, Vf = 0 m/s + (9.8 m/s^2)(2.36 s) = approx. 23.1 m/s.
C.) At what angle below the horizontal does it land? Answer in units of degrees
Now, let's dive into the angle at which the stone lands. We can find it using some trigonometry. The horizontal velocity of the stone remains constant throughout its flight, so we can use the horizontal distance and time to calculate it:
Horizontal Velocity = Distance / Time
Plugging in the values, Horizontal Velocity = 25 m / 2.36 s = approx. 10.59 m/s.
Now, let's find the vertical velocity at impact. The vertical velocity is influenced by gravity, so we can use the equation:
Vertical Velocity = (Acceleration due to gravity) x (Time)
Plugging in the values, Vertical Velocity = (9.8 m/s^2) x (2.36 s) = approx. 23.1 m/s.
To find the angle below the horizontal, we can use the tangent function:
Tangent of Angle = Vertical Velocity / Horizontal Velocity
Plugging in the values, Tangent of Angle = 23.1 m/s / 10.59 m/s.
Now, let's find the angle. Using an inverse tangent (arctan) function:
Angle = arctan(Tangent of Angle)
Calculating it gives us the answer: Angle ≈ 65.1 degrees.
So, the stone lands at an angle of approximately 65.1 degrees below the horizontal. It's quite a stylish way to make a splash on the beach!
To solve these questions, we can use the equations of motion for projectile motion. The horizontal and vertical motions are independent of each other, so we can analyze them separately.
Given:
Initial horizontal speed (vx) = 10.6 m/s
Vertical displacement (h) = -25 m (negative because the stone is falling downward)
Acceleration due to gravity (g) = 9.8 m/s^2
A.) To determine the time it takes for the stone to strike the beach, we will use the equation:
h = (1/2)gt^2
where t is the time taken.
Rearranging the equation, we have:
t = sqrt(2h / g)
Substituting the given values:
t = sqrt(2(-25) / 9.8)
t = sqrt(-50 / 9.8)
t = sqrt(-5.1)
However, we cannot take the square root of a negative number since it has no real solution. This means the stone will never strike the beach below the cliff.
B.) As the stone does not strike the beach, there is no speed to determine at impact.
C.) Similarly, since the stone does not reach the beach, there is no angle below the horizontal at which it lands.
To find the solutions to these questions, we need to break down the problem and apply the appropriate formulas.
A.) To find the time it takes for the stone to strike the beach below the cliff, we need to determine the time it takes for the stone to fall vertically. We can use the formula:
h = (1/2) * g * t^2
Where:
h = height (25 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the formula gives us:
t = sqrt(2h/g)
Substituting the given values, we have:
t = sqrt(2 * 25 / 9.8) ≈ 2.54 seconds
Therefore, the stone takes approximately 2.54 seconds to strike the beach below the cliff.
B.) To find the speed of the stone at impact, we need to use the horizontal velocity since the stone is thrown horizontally. The horizontal velocity remains constant throughout the motion. The formula to calculate the horizontal velocity is:
v = d / t
Where:
v = velocity (10.6 m/s)
d = horizontal distance covered by the stone (which is the time multiplied by the horizontal velocity)
t = time (2.54 seconds)
Rearranging the formula gives us:
d = v * t
Substituting the given values, we have:
d = 10.6 * 2.54 ≈ 26.924 meters
Therefore, the horizontal distance covered by the stone is approximately 26.924 meters.
C.) To find the angle below the horizontal at which the stone lands, we can use the tangent of the angle. The formula is:
tan(θ) = h / d
Where:
θ = angle below the horizontal
h = vertical distance (25 m)
d = horizontal distance (26.924 m)
Rearranging the formula gives us:
θ = arctan(h / d)
Substituting the given values, we have:
θ = arctan(25 / 26.924) ≈ 42.9 degrees
Therefore, the stone lands at an angle of approximately 42.9 degrees below the horizontal.
A. d = Vo*t + 0.5g*t^2 = 25 m.
0 + 4.9t^2 = 25
t^2 = 5.10
Tf = 2.26 s. = Fall time. = Time in air.
B. V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*25.
V^2 = 490
V = 22.1 m/s.