a rock is thrown horizontally with a speed of 17 m/s from a vertical cliff of height 34 m.
A- I solved for t=2.63
B-How far will it land from the base of the cliff?
im using x=xo+vox*t
i get -6.46 and it is wrong. What am I doing wrong?
C-What is the velocity (magnitude and direction counterclockwise from the +x-axis, which is the initial horizontal direction in which the rock was thrown) of the rock just before it hits the ground?
Magnitude m/s
Direction °
what would i use for a formula?
A- Your calculation for time, t=2.63 seconds, is correct.
B- To find how far the rock will land from the base of the cliff, you need to use the formula for horizontal displacement x = xo + vox*t. However, it seems like you might have made a mistake in the calculation. Let's go through it step by step.
Given:
Initial horizontal velocity, vox = 17 m/s
Time, t = 2.63 s
To find the horizontal displacement, x:
x = xo + vox*t
Since the rock is thrown horizontally, there is no initial horizontal displacement, so xo = 0.
x = 0 + 17 * 2.63
x = 44.71 m
Therefore, the rock will land approximately 44.71 meters from the base of the cliff.
C- To find the velocity of the rock just before it hits the ground, you can use the formula for vertical velocity, which is vy = voy + a*t. Here, voy is the initial vertical velocity, and a is the acceleration due to gravity (-9.8 m/s^2).
Given:
Initial vertical velocity, voy = 0 (since the rock is thrown horizontally without any upward or downward component)
Acceleration due to gravity, a = -9.8 m/s^2
Time, t = 2.63 s
To find the vertical velocity, vy:
vy = voy + a*t
vy = 0 + (-9.8) * 2.63
vy = -25.774 m/s (rounded to three decimal places)
The magnitude of the velocity just before the rock hits the ground is 25.774 m/s (rounded to three decimal places).
Since the rock is thrown horizontally, the direction is along the negative y-axis, counterclockwise from the +x-axis. In this case, the direction will be 90 degrees counterclockwise from the +x-axis.
Therefore, the velocity of the rock just before it hits the ground has a magnitude of 25.774 m/s and is directed 90 degrees counterclockwise from the +x-axis.
A- To solve for time (t), you can use the equation: h = gt^2/2, where h is the height and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the height (h) is 34 m.
So, 34 = (9.8 * t^2) / 2
Multiplying both sides by 2 and dividing by 9.8, we get:
t^2 = (34 * 2) / 9.8
t^2 = 69.39 / 9.8
t^2 ≈ 7.08
Taking the square root of both sides, we can find t:
t ≈ √7.08 ≈ 2.66 seconds (approximately)
So, t = 2.66 seconds (approximately).
It seems like you made a calculation mistake in solving for time (t). You mentioned t = 2.63, but the correct value is around 2.66 seconds.
B- To find how far the rock will land from the base of the cliff, you can use the equation: x = x0 + v0x * t, where x is the distance, x0 is the initial position (which is usually taken as 0), v0x is the initial horizontal velocity, and t is the time.
In this case, the initial horizontal velocity (v0x) is 17 m/s, and the time (t) is approximately 2.66 seconds.
So, x = 0 + 17 * 2.66
x ≈ 45.22 meters (approximately)
Therefore, the rock will land around 45.22 meters away from the base of the cliff.
C- Just before it hits the ground, the vertical velocity of the rock will be equal to the initial upward velocity (as the only force acting on the rock is gravity). The horizontal velocity will remain constant throughout the motion, as there is no horizontal force acting on the rock.
The magnitude of the horizontal velocity (v0x) is 17 m/s, which is the same velocity with which the rock was thrown.
The magnitude of the vertical velocity just before hitting the ground can be found using the equation: v = gt. In this case, g is approximately 9.8 m/s^2 (acceleration due to gravity) and t is approximately 2.66 seconds.
So, v = 9.8 * 2.66 ≈ 26.12 m/s (approximately)
Therefore, the magnitude of the velocity just before hitting the ground is approximately 26.12 m/s.
Since the rock was initially thrown horizontally, the direction of this velocity will be purely vertical (downward).
Hence, the direction counterclockwise from the +x-axis will be 270° (or -90°).
Magnitude: 26.12 m/s
Direction: -90° (counterclockwise from the +x-axis)
A- To solve for the time (t), you can use the equation of motion for vertical motion, which is given as:
y = yo + voy*t - (1/2)gt^2
In this case, you know that the initial vertical position (yo) is 34 m (since the rock is thrown from a vertical cliff), the initial vertical velocity (voy) is 0 m/s (since the rock is thrown horizontally), and the acceleration due to gravity (g) is approximately 9.8 m/s^2.
Plugging in these values, the equation becomes:
34 = 0*t - (1/2)*9.8*t^2
Simplifying this equation, we get:
4.9*t^2 = 34
Dividing both sides by 4.9, we get:
t^2 = 6.9387755
Taking the square root of both sides, we find:
t ≈ 2.63 seconds (rounded to two decimal places)
Therefore, your calculation for t is correct.
B- To find the horizontal distance (x) the rock travels, you can use the equation of motion for horizontal motion, which is given as:
x = xo + vox * t
In this case, the initial horizontal position (xo) is 0 m (since the rock is thrown horizontally), the initial horizontal velocity (vox) is 17 m/s (as given in the question), and the time (t) is 2.63 seconds (as you have calculated).
Substituting these values into the equation, we get:
x = 0 + 17 * 2.63
x ≈ 44.71 meters (rounded to two decimal places)
Therefore, the rock will land approximately 44.71 meters from the base of the cliff.
C- To find the magnitude and direction of the velocity just before the rock hits the ground, you can use the horizontal and vertical components of the velocity.
The magnitude of the velocity (v) can be found using the Pythagorean theorem:
v = sqrt((vox)^2 + (voy)^2)
In this case, the horizontal velocity (vox) is 17 m/s (as given in the question), and the vertical velocity (voy) is 0 m/s (since the rock is thrown horizontally). Therefore, the magnitude of the velocity is:
v = sqrt((17)^2 + (0)^2) = 17 m/s
The direction counterclockwise from the +x-axis can be found using:
θ = tan^(-1)(voy / vox)
In this case, since the vertical velocity (voy) is 0 m/s, the direction is directly along the +x-axis.
So, the magnitude of the velocity just before the rock hits the ground is 17 m/s, and the direction is along the +x-axis, which is the initial horizontal direction in which the rock was thrown.