A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 12.6 m/s. The cliff is 49.1 m above a flat horizontal beach as shown in the figure.
a.How long after being released does the stone strike the beach below the cliff?
b.With what speed does the stone land?
c.What is the angle of impact at landing?
Vertical problem:
v = a t
x = Xi + Vi t + .5 a t^2
here
x = 0
Xi = 49.1
Vi = 0
a = -9.8
0 = 49.1 + 0 t - 4.9 t^2
so
a) t = sqrt ( 49.1 m/4.9 m/s^2 )
b) v = -9.8 t
c)
u = 12.6
v = -9.8 t
tan(angle from horizontal) =(v/u)
To solve this problem, we can use the equations of motion in the horizontal and vertical directions. Let's break down each part.
a. How long after being released does the stone strike the beach below the cliff?
We know that the stone is thrown horizontally, which means its initial vertical velocity is 0 m/s. The only force acting on the stone vertically is gravity. We can use the equation of motion for vertical displacement:
Δy = V0yt + (1/2)gt^2
where:
Δy = vertical displacement (49.1 m, the height of the cliff)
V0y = initial vertical velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Plugging in the known values, the equation becomes:
49.1 = 0*t + (1/2)(-9.8)*t^2
Simplifying the equation:
49.1 = -4.9t^2
Rearranging the equation, we have:
t^2 = 10 (dividing both sides by -4.9)
Taking the square root of both sides, we get:
t ≈ √10
Therefore, the stone will strike the beach below the cliff approximately √10 seconds after being released.
b. With what speed does the stone land?
Since there is no horizontal acceleration, the horizontal component of the stone's velocity remains constant throughout its motion. We can use the formula for horizontal velocity:
Vx = V0x
where:
Vx = horizontal velocity
V0x = initial horizontal velocity
Given that the stone is thrown horizontally with a speed of 12.6 m/s, the horizontal velocity remains 12.6 m/s.
c. What is the angle of impact at landing?
The angle of impact can be found using trigonometry. We can calculate the tangent of the angle using the vertical and horizontal components of the stone's velocity:
tanθ = |Vy / Vx|
where:
θ = angle of impact
Vy = vertical component of the stone's velocity
Since the stone is thrown horizontally, Vy = 0 m/s.
tanθ = 0 / 12.6
tanθ = 0
To find the angle, we need to find the inverse tangent (or arctan) of 0, which is 0 degrees.
Therefore, the angle of impact at landing is 0 degrees (or horizontally).
To solve this problem, we can use the equations of motion for projectile motion. We'll start by analyzing the vertical and horizontal components separately.
a. How long after being released does the stone strike the beach below the cliff?
In the vertical direction, we can use the equation:
h = ut + (1/2)gt^2
Where:
h = height (distance) above the beach (49.1 m)
u = initial vertical velocity (0 m/s since the stone is thrown horizontally)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken (what we need to find)
Since the stone was thrown horizontally, the initial vertical velocity is zero. So, the equation simplifies to:
h = (1/2)gt^2
Substituting the given values:
49.1 = (1/2)(-9.8)t^2
t^2 = (2 * 49.1) / 9.8
t^2 = 100
t = sqrt(100)
t = 10 s
Therefore, it takes 10 seconds for the stone to strike the beach below the cliff.
b. With what speed does the stone land?
In the horizontal direction, the stone maintains a constant horizontal speed throughout its motion. Hence, the initial horizontal velocity is the same as the final horizontal velocity.
Given:
Initial horizontal velocity (u) = 12.6 m/s
Therefore, the stone lands with a horizontal velocity of 12.6 m/s.
c. What is the angle of impact at landing?
To find the angle of impact, we can use trigonometry. The horizontal velocity remains constant, but the vertical velocity changes due to gravity. The angle of impact (θ) can be calculated as:
θ = tan^(-1)(vertical velocity / horizontal velocity)
The vertical velocity at the time of landing can be found using the equation of motion:
v = u + gt
Substituting the given values:
v = 0 + (-9.8) * 10
v = -98 m/s (negative because it is directed downward)
Substituting the values into the angle equation:
θ = tan^(-1)(-98 / 12.6)
θ = -80.1 degrees
Therefore, the angle of impact at landing is approximately -80.1 degrees (negative sign indicates the downward direction).