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

[sunnyday]

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.

 

[PeroK]

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 $21$?

 

[ehild]

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.

 

[sunnyday]

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).

 

[PeroK]

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?

 

[ehild]

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?

 

[hadaidea]

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

 

[hadaidea]

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

 

[haruspex]

btw the answer is 27m

That's not what I get. Please post your working.
Btw, the thread is five years old.