a rock is thrown of a cliff with an horizontal velocity of 55.0 m/sec. it takes the rock 7.0 sec to hit the ground
a) what is the height of the cliff.
b)how far from the foot of the cliff does the rock hit the ground.
c) what is the vertical component of velocity with which its hits the ground.
d) what is the vertical component of velocity with which hits the ground.
a) There is an equation for distance fallen vs time that you should know.
y = (g/2) t^2
Use it to solve for t.
b) 55.0 t
c) g*t
d) same question as c). Same answer. You probably typed part of the question wrong.
To find the answers to the given questions, we can use the kinematic equations of motion. Let's break down each question and solve them one by one.
a) What is the height of the cliff?
To find the height of the cliff, we need to use the equation: h = (1/2) * g * t^2, where h is the height, g is the acceleration due to gravity, and t is the time taken to hit the ground. Since the horizontal velocity has no effect on the height, we can ignore it in this calculation.
Given:
Horizontal velocity, Vx = 55.0 m/s
Time taken, t = 7.0 s
The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Substituting the values into the equation, we have:
h = (1/2) * (9.8 m/s^2) * (7.0 s)^2
h = 0.5 * 9.8 m/s^2 * 49.0 s^2
h = 240.1 m
Therefore, the height of the cliff is approximately 240.1 meters.
b) How far from the foot of the cliff does the rock hit the ground?
To find the horizontal distance, we can use the equation: d = Vx * t, where d is the distance, Vx is the horizontal velocity, and t is the time taken.
Given:
Horizontal velocity, Vx = 55.0 m/s
Time taken, t = 7.0 s
Substituting the values into the equation, we have:
d = (55.0 m/s) * (7.0 s)
d = 385.0 m
Therefore, the rock will hit the ground approximately 385.0 meters away from the foot of the cliff.
c) What is the vertical component of velocity with which it hits the ground?
Since the rock falls vertically, we know that its vertical velocity component is affected by gravity. We can calculate the vertical velocity, Vy, using the equation: Vy = g * t, where Vy is the vertical velocity and g is the acceleration due to gravity.
Given:
Time taken, t = 7.0 s
Acceleration due to gravity, g = -9.8 m/s^2 (negative because it acts in the opposite direction of motion)
Substituting the values into the equation, we have:
Vy = (9.8 m/s^2) * (7.0 s)
Vy = 68.6 m/s
Therefore, the vertical component of velocity with which the rock hits the ground is approximately 68.6 m/s (downwards).
d) What is the vertical component of velocity with which it hits the ground?
It seems like the question in point d is the same as the one in point c. So, as mentioned above, the vertical component of velocity with which the rock hits the ground is approximately 68.6 m/s (downwards).