A stone was dropped off a cliff and hit the ground with a speed of 152 ft/s. What is the height of the cliff? (Use 32 ft/s2 for the acceleration due to gravity.)
Question
A stone was dropped off a cliff and hit the ground with a speed of 152 ft/s. What is the height of the cliff? (Use 32 ft/s2 for the acceleration due to gravity.)
Solution
To solve this problem, we can use the equation for the final velocity of an object in free fall, which is given by:
v_f^2 = v_i^2 + 2gh
where:
- v_f is the final velocity (152 ft/s in this case),
- v_i is the initial velocity (0 ft/s in this case, since the stone was dropped, not thrown),
- g is the acceleration due to gravity (32 ft/s^2), and
- h is the height we're trying to find.
Substituting the given values into the equation, we get:
(152 ft/s)^2 = (0 ft/s)^2 + 2*(32 ft/s^2)*h
Solving for h, we get:
h = ((152 ft/s)^2 - (0 ft/s)^2) / (2*(32 ft/s^2))
h = (23104 ft^2/s^2) / (64 ft/s^2)
h = 361 ft
So, the height of the cliff is 361 feet.
Similar Questions
When the stone hits the ground, its height is 0 ft. We can now use0 = s(4.75) = −16 2 + h.Solving for h we find that the height of the cliff ish = ft.
A hiker throws a stone from the upper edge of a vertical cliff. The stone's initial velocity is 25.0 m/s directed at 40.0° with the face of the cliff, as shown in the figure.The stone hits the ground 3.75 s after being thrown.How far from the foot of the cliff does the stone land?Ignore air friction and express your answer in meters.
A rock is thrown straight down with an initial velocity of 17.5 m/s from a cliff. What is the rock's displacement after 2.0 s? (Acceleration due to gravity is 9.80 m/s2.)Select one:a.55 mb.28 mc.49 md.64 m
A stone thrown vertically upward at12m/s. How high would the stone be when its speed is 8m/s?a.4.1mb.4.5mc.4.8md.4.9m
A 5 kg stone is dropped from a height of 2 m. What is the speed of the stone when it hits the floor? Consider g = 10 ms-2Question 2Select one:a.10 m/sb.12.6 m/sc.5 m/sd.2 m/se.6.3 m/s
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.