Question

A stone thrown upward from the top of a 320-foot cliff at 128 ft/sec eventually falls to the beach below.

a) How long does the stone take to reach its highest point?
b) What is its maximum height?
c) How long before the stone hits the beach?
d) What is the velocity of the stone on impact?

Answer 1

a) It takes the stone four seconds to reach its peak.. b) The maximum height of the stone is 256 feet. c) The stone takes 2 seconds to hit the beach. d) The velocity of the stone on impact is -64 ft/sec.

Step-by-step explanation:

a) How long does the stone take to reach its highest point?

To find out the time taken to reach the highest point, we need to use the equation for vertical motion under constant acceleration:

v = u + at

The initial velocity, u, is 128 ft/sec (since the stone is thrown upward), and the final velocity, v, is 0 ft/sec at the highest point (since the stone momentarily comes to rest). The acceleration, a, due to gravity is -32 ft/sec² (taking the downward direction as negative).

Plugging the values into the equation, we get:

0 = 128 - 32t

Simplifying the equation, we find that t = 4 seconds. As a result, it takes the stone 4 seconds to reach its peak. b) What is the tallest point on it? We can use the vertical displacement equation to get the maximum height:

s = ut + (1/2)at²

At the highest point, the final velocity is 0 ft/sec, so we can use the time taken to reach the highest point, t, which is 4 seconds. Plugging in the values, we find that s = 128(4) + (1/2)(-32)(4)² = 256 ft. Therefore, the maximum height of the stone is 256 feet.

c) How long before the stone hits the beach?

Since the stone was thrown from a height of 320 feet and the maximum height it reaches is 256 feet, the stone falls a distance of 320 - 256 = 64 feet to reach the beach. We can use the equation for vertical displacement again:

s = ut + (1/2)at²

The initial velocity, u, is 0 ft/sec (since the stone is falling) and the acceleration, a, due to gravity is -32 ft/sec² (taking the downward direction as negative). Plugging in the values, we find:

64 = 0t + (1/2)(-32)t²

Simplifying the equation, we get a quadratic equation t² = 4. The negative solution is not applicable in this context, so the positive solution is t = 2 seconds. Therefore, it takes 2 seconds for the stone to hit the beach.

d) What is the velocity of the stone on impact?

On impact, the stone has a final velocity, v. We can use the equation for vertical motion to find this:

v = u + at

The initial velocity, u, is 0 ft/sec (since the stone is falling) and the acceleration, a, due to gravity is -32 ft/sec² (taking the downward direction as negative). The time, t, is 2 seconds (as calculated in part c). Plugging in the values, we find that v = 0 - 32(2) = -64 ft/sec. Therefore, the velocity of the stone on impact is -64 ft/sec.

Answer 2

The stone takes 4 seconds to reach its highest point, reaches a maximum height of 256 feet, takes 2 seconds to hit the beach, and has a velocity of 64 ft/sec on impact.

Step-by-step explanation:

To solve this problem, we can use the equations of motion for projectile motion. Let's break down each part of the question:

(a) How long does the stone take to reach its highest point?

When the stone reaches its highest point, its vertical velocity will become zero. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial vertical velocity is 128 ft/sec and the acceleration is -32 ft/sec² (due to gravity). So, we have 0 = 128 - 32t. Solving for t, we get t = 4 seconds.

(b) What is its maximum height?

To find the maximum height, we can use the equation h = ut + 0.5at², where h is the height, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we have h = 128(4) + 0.5(-32)(4)². Simplifying, we get h = 256 feet.

(c) How long before the stone hits the beach?

When the stone hits the beach, its height will be 0. We can use the equation h = ut + 0.5at² and solve for t. Plugging in the values, we have 0 = 128t + 0.5(-32)t². Simplifying, we get 16t² - 32t = 0. Factoring out t, we have t(16t - 32) = 0. So either t = 0 (which is the initial time) or 16t - 32 = 0. Solving the second equation, we get t = 2 seconds. Therefore, the stone takes 2 seconds to hit the beach.

(d) What is the velocity of the stone on impact?

When the stone hits the beach, its velocity will be the final velocity. We can use the equation v = u + at to find it. Plugging in the values, we have v = 128 - 32(2). Simplifying, we get v = 64 ft/sec.