A person kicks a rock off a cliff horizontally with a speed of 20.0 m/s. It takes 7.0 seconds to hit the ground, find:
a. height of the cliff
b. final vertical velocity
c. horizontal range
This problem involved half projectile.
initial velocity, vo = 20 m/s
time of flight, t = 7 s
(a) Simply use the formula to get the height, h:
h = vo*t - (1/2)gt^2
(b) To get the final vertical velocity or terminal velocity (vf), use the formula:
(vf)^2 - (vo)^2 = 2gh
(c) Use the formula find the horizontal distance traveled, R:
R = vo * cos(θ) * t
But since the angle involved with respect to horizontal is zero, and cos(0) = 1, we have
R = vo * t
Hope this helps~ `u`
But what's g?
a. Ah, the classic rock-kicking-off-a-cliff scenario! To find the height of the cliff, we can use the equation of motion:
\(h = \frac{1}{2}gt^2\), where \(h\) is the height, \(g\) is the acceleration due to gravity, and \(t\) is the time taken.
Using \(g = 9.8 m/s^2\) (approximately), we can plug in the values to get:
\(h = \frac{1}{2} \times 9.8 \times (7.0^2)\)
Calculating... and we get the answer: the height of the cliff is approximately 240.1 meters.
b. Now, for the final vertical velocity. When the rock hits the ground, its vertical velocity will be the negative of its initial vertical velocity. Since the rock was kicked horizontally, its initial vertical velocity is 0 m/s. Therefore, the final vertical velocity is -0 m/s. So, if you ever meet a rock with zero vertical velocity, be sure to give it a high five... or a low five in this case!
c. Lastly, let's find the horizontal range. The horizontal distance covered by an object in projectile motion can be calculated using the formula: \(R = V_{\text{horizontal}} \times t\), where \(R\) is the range and \(V_{\text{horizontal}}\) is the horizontal component of the initial velocity.
Since the rock was kicked horizontally, the horizontal component of the initial velocity is equal to the initial speed: 20.0 m/s. Plugging in the values, we have:
\(R = 20.0 \times 7.0\)
Calculating... and we find the horizontal range is approximately 140.0 meters. So, if you're ever looking for a rock you kicked off a cliff, start your search about 140.0 meters away from the base!
To find the height of the cliff (a), we need to use the equation of motion that relates time, initial velocity, acceleration, and displacement:
y = y0 + v0y*t + (1/2)*a*t^2
Where:
y = final vertical displacement (height of the cliff)
y0 = initial vertical displacement (0, since we are starting at ground level)
v0y = initial vertical velocity (0, since the rock is kicked horizontally)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (7.0 seconds)
Plugging in the values, we can solve for y:
y = 0 + 0*7.0 + (1/2)*(-9.8)*(7.0)^2
y = 0 - 0 - 33.6
y = -33.6 meters
Since the rock is falling downward, the negative sign indicates the direction of displacement. Thus, the height of the cliff is 33.6 meters.
To find the final vertical velocity (b), we use the equation:
v = v0 + a*t
Where:
v = final vertical velocity
v0 = initial vertical velocity (0, since the rock is kicked horizontally)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (7.0 seconds)
Plugging in the values, we can solve for v:
v = 0 + (-9.8)*7.0
v = -68.6 m/s
Again, the negative sign indicates the direction of motion. The final vertical velocity of the rock is -68.6 m/s.
To find the horizontal range (c), we multiply the initial horizontal velocity with the time taken:
Range = v0x * t
Where:
v0x = initial horizontal velocity (20.0 m/s)
t = time (7.0 seconds)
Plugging in the values, we can solve for the range:
Range = 20.0 * 7.0
Range = 140.0 meters
Therefore, the horizontal range of the rock is 140.0 meters.
To find the answers to the given questions, we can break down the problem and apply relevant physics equations. Let's start with the given information:
Initial horizontal velocity (Vx) = 20.0 m/s
Time of flight (t) = 7.0 s
a. To find the height of the cliff (h), we need to use the kinematic equation for vertical motion:
h = Vyi * t + (1/2) * a * t^2
Since the rock is kicked horizontally, its vertical initial velocity (Vyi) is zero, and the acceleration due to gravity (a) is approximately 9.8 m/s^2. Plugging in the values, we get:
h = 0 * 7.0 + (1/2) * 9.8 * (7.0)^2
h = 0 + (1/2) * 9.8 * 49.0
h = 0 + 4.9 * 49.0
h = 240.1 meters
Therefore, the height of the cliff is 240.1 meters.
b. To find the final vertical velocity (Vyf) when it hits the ground, we can use the kinematic equation:
Vyf = Vyi + a * t
Since the initial vertical velocity (Vyi) is zero, we can simplify the equation to:
Vyf = a * t
Vyf = 9.8 * 7.0
Vyf = 68.6 m/s
Hence, the final vertical velocity of the rock when it hits the ground is 68.6 m/s.
c. To find the horizontal range (R), we can use the equation:
R = Vx * t
Plugging in the given values, we get:
R = 20.0 * 7.0
R = 140.0 meters
Therefore, the horizontal range of the rock is 140.0 meters.