REPOST PLEASE HELP 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?
Your time is correct, but they want the horizontal distance from the base opf the cliff. How can that be negative?
X = 17 m/s*2.63 = 44.8 m
Your formula for x is crrect, but xo = 0 and I don't see how you came up with -6.46.
For the speed when it hits the ground, you can use conservation of energy.
[Vfinal^2 - Vinitial^2]/2 = g H
For the direction,
Vinitial/Vfinal = cos^-1 theta
Vinitial = Vx (during flight)= 17 m/s
A) To solve for the time it takes for the rock to reach the ground, you can use the formula for free-fall motion: h = (1/2)gt^2, where h is the height of the cliff (34 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes for the rock to reach the ground. Rearranging the equation, t = sqrt(2h/g).
t = sqrt(2 * 34 m / 9.8 m/s^2)
t = sqrt(68/9.8)
t ≈ 2.63 seconds
It seems you have solved for the correct value of t.
B) To determine how far the rock will land from the base of the cliff, you can use the equation x = xo + voxt, where x is the horizontal distance traveled, xo is the initial position (0 m since the rock was thrown horizontally from the cliff), vox is the initial horizontal velocity (17 m/s), and t is the time it takes for the rock to reach the ground.
x = 0 + (17 m/s) * (2.63 s)
x ≈ 44.71 meters
So, the rock will land approximately 44.71 meters from the base of the cliff.
It looks like you made a calculation mistake, resulting in an incorrect value. Double-check your calculations to see if you made any errors.
C) To find the velocity (magnitude and direction counterclockwise from the +x-axis) of the rock just before it hits the ground, you can use the equation v = vox + at, where v is the final velocity (which is the velocity just before it hits the ground), vox is the initial horizontal velocity (17 m/s), a is the acceleration due to gravity (-9.8 m/s^2 since it acts in the opposite direction of the initial velocity), and t is the time it takes for the rock to reach the ground.
v = (17 m/s) + (-9.8 m/s^2) * (2.63 s)
v ≈ 17 m/s - 25.74 m/s
v ≈ -8.74 m/s
Since the velocity is negative, it indicates that the rock is moving in the opposite direction of the initial horizontal velocity (counterclockwise from the +x-axis). The magnitude of the velocity is 8.74 m/s.
Therefore, the velocity of the rock just before it hits the ground has a magnitude of approximately 8.74 m/s and is in the counterclockwise direction from the +x-axis.