A ball is dropped from the top of a 55.0 m high cliff

In summary: Why bump it now?In summary, a ball is dropped from the top of a 55.0 m-high cliff and a stone is thrown straight up from the bottom of the cliff at the same time. They collide part way up and the question is asking for the distance from the base of the cliff where this happens. Using the equations for distance and time, we can solve for the time it takes for the stone and ball to collide. Plugging this time into the equations for distance, we get two answers: 33.6109 m for the ball and 21.389 m for the stone. However, the correct answer is not 27 m as stated, but rather 27.389 m. It appears that the
  • #1
sunnyday
6
0

Homework Statement


A ball is dropped from the top of a 55.0 m -high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 21.0 m/s . The stone and ball collide part way up.

How far above the base of the cliff does this happen?

Homework Equations


d = vt+1/2at^2
d1 + d2 = 55

The Attempt at a Solution


Ball[/B]
v = 0
d = 1/2*9.8t^2

Stone
v = 21
d = 21t-1/2*9.8t^2 (I made it -9.8 because the stone is going up so it's fighting gravity, I hope I'm right)

Ball + Stone = 55
1/2*9.8t^2 + 21t-1/2*9.8t^2 = 55
21t = 55
t = 2.619

Stone:
21(2.619)-(1/2)(9.8)(2.619)^2 = 21.389

Ball:
(1/2)(9.8)(2.619)^2 = 33.6109

I don't understand what I'm supposed to do next.
 
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  • #2
sunnyday said:

Homework Statement


A ball is dropped from the top of a 55.0 m -high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 21.0 m/s . The stone and ball collide part way up.

How far above the base of the cliff does this happen?

Homework Equations


d = vt+1/2at^2
d1 + d2 = 55

The Attempt at a Solution


Ball[/B]
v = 0
d = 1/2*9.8t^2

Stone
v = 21
d = 21t-1/2*9.8t^2 (I made it -9.8 because the stone is going up so it's fighting gravity, I hope I'm right)

Ball + Stone = 55
1/2*9.8t^2 + 21t-1/2*9.8t^2 = 55
21t = 55
t = 2.619

Stone:
21(2.619)-(1/2)(9.8)(2.619)^2 = 21.389

Ball:
(1/2)(9.8)(2.619)^2 = 33.6109

I don't understand what I'm supposed to do next.

What do you mean next? You've got the answer, surely?

Note that you need to add units to your working, e.g. ##t = 2.619s##.

An an aside: the stone starts off at ##21m/s## and the initial distance between the stone and the ball is ##55m##. What do you get if you divide ##55## by ##21##?
 
  • #3
Stone:
21(2.619)-(1/2)(9.8)(2.619)^2 = 21.389

Ball:
(1/2)(9.8)(2.619)^2 = 33.6109

I don't understand what I'm supposed to do next.[/QUOTE]
You have to answer the question: How far above the base of the cliff does this happen?
You got two numbers, explain what they mean. First, add units, as @PeroK said.
Clarify, at what distance from the base of the cliff the ball and stone collide.
 
  • #4
PeroK said:
What do you mean next? You've got the answer, surely?

Note that you need to add units to your working, e.g. ##t = 2.619s##.

An an aside: the stone starts off at ##21m/s## and the initial distance between the stone and the ball is ##55m##. What do you get if you divide ##55## by ##21##?

When I divided 55/21, I got 2.619 (the same number as I got for seconds). I got 2 answers, the displacement for the ball (33.6109m) and the displacement for the stone (21.389m).
 
  • #5
sunnyday said:
When I divided 55/21, I got 2.619 (the same number as I got for seconds). I got 2 answers, the displacement for the ball (33.6109m) and the displacement for the stone (21.389m).

How far about the cliff is the collision? Hint: think about the stone.

Is ##55/21 = 2.619## a coincidence?
 
  • #6
sunnyday said:
I got 2 answers, the displacement for the ball (33.6109m) and the displacement for the stone (21.389m).
Answer the question: How far above the base of the cliff does the collision happen?
 
  • #7
sunnyday said:

Homework Statement


A ball is dropped from the top of a 55.0 m -high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 21.0 m/s . The stone and ball collide part way up.

How far above the base of the cliff does this happen?

Homework Equations


d = vt+1/2at^2
d1 + d2 = 55

The Attempt at a Solution


Ball[/B]
v = 0
d = 1/2*9.8t^2

Stone
v = 21
d = 21t-1/2*9.8t^2 (I made it -9.8 because the stone is going up so it's fighting gravity, I hope I'm right)

Ball + Stone = 55
1/2*9.8t^2 + 21t-1/2*9.8t^2 = 55
21t = 55
t = 2.619

Stone:
21(2.619)-(1/2)(9.8)(2.619)^2 = 21.389

Ball:
(1/2)(9.8)(2.619)^2 = 33.6109

I don't understand what I'm supposed to do next.
btw the answer is 27m
 
  • #8
sunnyday said:

Homework Statement


A ball is dropped from the top of a 55.0 m -high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 21.0 m/s . The stone and ball collide part way up.

How far above the base of the cliff does this happen?

Homework Equations


d = vt+1/2at^2
d1 + d2 = 55

The Attempt at a Solution


Ball[/B]
v = 0
d = 1/2*9.8t^2

Stone
v = 21
d = 21t-1/2*9.8t^2 (I made it -9.8 because the stone is going up so it's fighting gravity, I hope I'm right)

Ball + Stone = 55
1/2*9.8t^2 + 21t-1/2*9.8t^2 = 55
21t = 55
t = 2.619

Stone:
21(2.619)-(1/2)(9.8)(2.619)^2 = 21.389

Ball:
(1/2)(9.8)(2.619)^2 = 33.6109

I don't understand what I'm supposed to do next.
btw the answer is 27m
 
  • #9
hadaidea said:
btw the answer is 27m
That's not what I get. Please post your working.
Btw, the thread is five years old.
 

1. How long does it take for the ball to reach the ground?

The time it takes for the ball to reach the ground can be calculated using the equation t = √(2h/g), where t is the time in seconds, h is the initial height in meters, and g is the acceleration due to gravity (9.8 m/s²). Substituting the given values, we get t = √(2(55.0)/9.8) = 3.16 seconds.

2. What is the velocity of the ball just before it reaches the ground?

The velocity of the ball just before it reaches the ground can be calculated using the equation v = √(2gh), where v is the velocity in meters per second, g is the acceleration due to gravity, and h is the initial height. Substituting the given values, we get v = √(2(9.8)(55.0)) = 42.43 m/s.

3. How far does the ball travel horizontally before hitting the ground?

The horizontal distance traveled by the ball can be calculated using the equation d = v*t, where d is the distance in meters, v is the horizontal velocity, and t is the time in seconds. Since there is no initial horizontal velocity (assuming the ball is dropped straight down), we can substitute the value of the time calculated in question 1 to get d = (0)(3.16) = 0 meters. This means that the ball does not travel horizontally before hitting the ground.

4. What is the acceleration of the ball during its fall?

The acceleration of the ball during its fall is constant and equal to the acceleration due to gravity, which is 9.8 m/s². This means that the ball's velocity increases by 9.8 meters per second every second it falls.

5. What is the potential energy of the ball at the top of the cliff and the kinetic energy just before it reaches the ground?

At the top of the cliff, the ball has potential energy equal to mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the cliff. Assuming a mass of 1 kilogram for the ball, we can calculate the potential energy as (1)(9.8)(55.0) = 539 joules. Just before it reaches the ground, all of the potential energy has been converted to kinetic energy, so the kinetic energy is also 539 joules.

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