Dropped ball Kinematics motion

In summary, to throw the ball downward so that it and the dropped ball hit the ground at the same time, take the square root of a negative number and use g to represent the speed.
  • #1
negation
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Homework Statement



You're atop a building of height, h, and a friend is poised to drop a ball from a window at h/2. Find an expression for the speed at which you should simultaneously throw a ball downward, so the two hit the ground at the same time.

Homework Equations



None

The Attempt at a Solution



1)H = vit + 0.5gt^2
t1 = SQRT[(H - vit)/-4.9ms^-2]

2)H/2 = vit + 0.5gt^2
t2 = SQRT[(H/2-vit)/-4.9ms6-2]
t2 is the time taken for ball to fall from H/2 to ground

t2=t1
so,
set t1= SQRT[(H - vit)/-4.9ms^-2] to be equals to t2
therefore,
SQRT[(H/2-vit)/-4.9ms6-2] = SQRT[(H - vit)/-4.9ms^-2]

vi = [H - 4.9ms^-2(h/2-vit)/(-4.9ms^-2)]/t

Could someone give me a let up?
 
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  • #2
Hello.

Note that one of the balls is dropped. For this ball, try to find an expression for the time it takes the ball to reach the ground in terms of H and g.
 
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  • #3
TSny said:
Hello.

Note that one of the balls is dropped. For this ball, try to find an expression for the time it takes the ball to reach the ground in terms of H and g.

If the ball is dropped from H/2, then the assumption vi=0 can be made.
Hence, H/2 = -4.9ms^-2t^2
t = SQRT[(H/2)/-4.9ms^-2]

As for ball dropped from H:

H = vit + 0.5at^2
H = t(vi + 0.5at) = vi + 0.5at
t = [H -vi]/05a = [H - vi]/-4.9ms^-2
 
  • #4
negation said:
If the ball is dropped from H/2, then the assumption vi=0 can be made.
Hence, H/2 = -4.9ms^-2t^2
t = SQRT[(H/2)/-4.9ms^-2]

OK, but it looks like you've run into a sign problem. You have an expression where you're taking the square root of a negative number.

I suggest that you express your results in terms of the symbol g and not substitute a numerical value for g. (Since a value for H is not given, you won't be able to find a numerical answer to the question.)

As for ball dropped from H:

H = vit + 0.5at^2
H = t(vi + 0.5at) = vi + 0.5at

You can't just drop out the factor of t

Is there any relationship between the t in this equation and the t you got for the ball that was dropped from H/2?
 
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  • #5
TSny said:
OK, but it looks like you've run into a sign problem. You have an expression where you're taking the square root of a negative number.

I suggest that you express your results in terms of the symbol g and not substitute a numerical value for g. (Since a value for H is not given, you won't be able to find a numerical answer to the question.)
H/2 = 0.5gt^2
t = SQRT[H/g]
TSny said:
You can't just drop out the factor of t

Is there any relationship between the t in this equation and the t you got for the ball that was dropped from H/2?

What a terrible elementary blunder.

It appears logical to substitute t = SQRT[H/g] into H = vit + 0.5gt^2. Then isolate vi from the other variables.
The ball thrown from H and the ball dropped from H/2 must both touch the ground at t = SQRT[H/g]
From this, it can be deduced that the distance H = vit + 0.5gt^2 must be achieved in t = SQRT[H/g]
 
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  • #6
negation said:
H/2 = 0.5gt^2
t = SQRT[H/g]

It appears logical to substitute t = SQRT[H/g] into H = vit + 0.5gt^2. Then isolate vi from the other variables.
The ball thrown from H and the ball dropped from H/2 must both touch the ground at t = SQRT[H/g]
From this, it can be deduced that the distance H = vit + 0.5gt^2 must be achieved in t = SQRT[H/g]

Sounds good!
 
  • #7
TSny said:
Sounds good!
Solved
 
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Related to Dropped ball Kinematics motion

1. What is dropped ball kinematics motion?

Dropped ball kinematics motion refers to the movement of an object (such as a ball) that is released from a certain height and falls towards the ground due to the force of gravity. It follows the laws of kinematics, which describe the motion of objects without considering the forces that cause the motion.

2. How is the velocity of a dropped ball calculated?

The velocity of a dropped ball can be calculated using the formula v = √(2gh), where v is the velocity in meters per second (m/s), g is the acceleration due to gravity (9.8 m/s²), and h is the height of the ball in meters.

3. What is the acceleration of a dropped ball?

The acceleration of a dropped ball is constant and equal to the acceleration due to gravity, which is approximately 9.8 m/s². This means that the ball's velocity increases by 9.8 m/s every second as it falls towards the ground.

4. How does air resistance affect the motion of a dropped ball?

Air resistance, also known as drag force, can slow down the motion of a dropped ball. This force increases as the ball falls faster, and eventually, it will reach a terminal velocity where the drag force is equal to the force of gravity, and the ball will stop accelerating and fall at a constant speed.

5. Can the height from which a ball is dropped affect its motion?

Yes, the height from which a ball is dropped can affect its motion. The higher the ball is dropped from, the longer it will take to reach the ground and the higher its final velocity will be. This is because it has more time to accelerate due to gravity. Additionally, the height can also affect the amount of air resistance the ball experiences, which can alter its motion.

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