A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 16.0 m/s. The cliff is h = 25.0 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
a) how long does it take something to fall 25m?
hf=hi+vi*t-4.9t^2
hf=0, hi=25, vi=0 (vertical velocity), solve for time t.
speed: v=sqrt(vhorital^2+vf'^2)
where vf'=final vertical speed
= -9.8*timefalling
angle, measured from the horizontal, = arc (vf'/vh)
for the first part i got 2.26 seconds,
but for the: WHAT SPEED AND ANGLE OF IMPACT DOES THE STONE LAND?
IM NOT GETTING THE RIGHT ANSWER FOR THE SPEED WITH THE CALCULATION IM GETTING -22.2M/S WHICH IS WRONG. AND SO IS THE ANGLE.
To find the time it takes for the stone to strike the beach, we can use the formula for the time of flight of a projectile. The horizontal motion of the stone is independent of its vertical motion, so we can analyze the two motions separately.
1. First, let's find the time it takes for the stone to fall vertically from the cliff to the beach. We can use the equation of motion for vertical motion:
h = (1/2) * g * t^2
where h is the height of the cliff above the beach (25.0 m) and g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation to solve for t, we get:
t = sqrt(2h/g)
Substituting the given values, we have:
t = sqrt(2 * 25.0 / 9.8)
= sqrt(50 / 9.8)
= 3.19 s (rounded to two decimal places)
So, the time it takes for the stone to fall vertically to the beach is 3.19 seconds.
2. Now, let's find the time it takes for the stone to travel horizontally across the beach. Since the stone is thrown horizontally, its initial vertical velocity is zero. The only force acting on the stone horizontally is its initial velocity of 16.0 m/s.
The horizontal distance traveled (x) can be given by the equation:
x = v * t
where v is the horizontal velocity of the stone, and t is the time it takes to fall vertically.
Substituting the given values, we have:
x = 16.0 * 3.19
= 51.04 m (rounded to two decimal places)
So, the stone will strike the beach at a horizontal distance of 51.04 meters from the cliff.
3. To find the speed and angle of impact at which the stone lands, we can use trigonometry. The horizontal speed remains constant at 16.0 m/s, but the vertical speed changes due to the acceleration due to gravity.
The vertical velocity at impact (vf) can be found using the equation for vertical motion:
vf = gt
where g is the acceleration due to gravity and t is the time it takes to fall vertically.
Substituting the given values, we have:
vf = 9.8 * 3.19
= 31.28 m/s (rounded to two decimal places)
The speed of impact (v) can be found using the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
where vx is the horizontal velocity (16.0 m/s) and vy is the vertical velocity at impact (31.28 m/s).
Substituting the given values, we have:
v = sqrt(16.0^2 + 31.28^2)
= sqrt(256 + 979.5584)
= sqrt(1235.5584)
= 35.15 m/s (rounded to two decimal places)
The angle of impact (θ) can be found using the equation:
θ = arctan(vy / vx)
Substituting the given values, we have:
θ = arctan(31.28 / 16.0)
= arctan(1.955)
≈ 63.43 degrees (rounded to two decimal places)
So, the stone will land on the beach with a speed of approximately 35.15 m/s and an angle of impact of approximately 63.43 degrees below the horizontal.
To determine how long after being released the stone strikes the beach below the cliff, we can use the equation of motion in the vertical direction:
h = (1/2) * g * t^2
Where:
- h is the height of the cliff (25.0 m)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time taken for the stone to strike the beach below the cliff (unknown)
Rearranging the equation to solve for t:
t^2 = (2 * h) / g
t = sqrt((2 * h) / g)
Substituting the given values:
t = sqrt((2 * 25.0) / 9.8)
= sqrt(50.0 / 9.8)
= sqrt(5.10)
Therefore, the stone takes approximately 2.26 seconds to strike the beach below the cliff.
To determine the speed of impact, we need to find the horizontal distance traveled by the stone in 2.26 seconds.
Since the stone is thrown horizontally, it has a constant horizontal velocity of 16.0 m/s. We can calculate the horizontal distance using the formula:
d = v * t
Where:
- d is the horizontal distance traveled by the stone (unknown)
- v is the horizontal velocity (16.0 m/s)
- t is the time taken for the stone to strike the beach (2.26 seconds)
Substituting the values:
d = 16.0 * 2.26
Therefore, the horizontal distance traveled by the stone is approximately 36.16 meters.
Now, to find the speed and angle of impact, we can use basic trigonometry.
The speed of impact can be calculated using the horizontal and vertical distances traveled by the stone. We already know the horizontal distance is 36.16 meters, and the vertical distance is 25.0 meters (the height of the cliff).
The speed of impact can be found using the formula:
V = sqrt(Vx^2 + Vy^2)
Where:
- V is the speed of impact (unknown)
- Vx is the horizontal velocity (16.0 m/s)
- Vy is the vertical velocity (unknown)
To find Vy, we can use the equation:
Vy = g * t
Substituting the values:
Vy = 9.8 * 2.26
Now, we can substitute the values back into the speed of impact formula:
V = sqrt((16.0)^2 + (9.8 * 2.26)^2)
Finally, to find the angle of impact below the horizontal, we can use the equation:
θ = atan(Vy / Vx)
θ = atan((9.8 * 2.26) / 16.0)
Therefore, the stone lands with a speed of approximately ___ m/s and an angle of approximately ___ degrees below the horizontal.