Body thrown with v0 - find v0 that doubles prev max height

[physicsnnewbie]

Homework Statement

A body is thrown into the air with an initial velocity of v₀. What initial velocity is required to double the maximum height previously attained?

Homework Equations

The Attempt at a Solution

I found the max height v₀²/64 at v₀ by solving for 't' in the velocity equation when 'v' = 0, and subbing that for 't' in the height equation and solving. I then multiplied this by 2 and set it to the height and tried to solve for v₀ and ended up with v₀² = -512t² + 32tv₀. I then solved for v₀ and ended up with v₀ = 16t. Do I need to somehow find 't' or did i stray off course somewhere and end up with the wrong answer?

 

[tiny-tim]

hi physicsnnewbie!

wouldn't it be easier to use the other constant acceleration equation, v_f² = v_i² + 2as ?

 

[Beaker87]

Homework Statement

A body is thrown into the air with an initial velocity of v₀. What initial velocity is required to double the maximum height previously attained?

Homework Equations

The Attempt at a Solution

I found the max height v₀²/64 at v₀ by solving for 't' in the velocity equation when 'v' = 0, and subbing that for 't' in the height equation and solving.

You were good up until here. Now, all you need to do is figure out what number you need to multiply v₀ by in order to make the maximum height double.

in other words: find a constant C such that (Cv₀)²/64=2(v₀²/64)

 

[physicsnnewbie]

tim, I tried the formula you mentioned but couldn't obtain the correct answer. Maybe I did something wrong. Beaker, it took me a while to figure out your equation, because i kept reading it as CV₀²/64 = 2V₀²/64 without paying enough attention to the parenthesis and i was thinking doesn't c just equal 2? :p.

 

[Apphysicist]

Tim's method also works, and may be of greater use to you in the future rather than the perhaps easier yet less universal method presented instead.

Under your initial condition of v₀ yielding a height, h, and noting that at the maximum height, v_f=0, you have v₀^2=2*a*h (because you assume the displacement, h up, is in the positive direction, a is then negative). Then in the second case, you have a new velocity, C*v₀ and a height 2*h. Working with the same idea and using the relationship you established in the first case, you can solve that as well.

 

[SammyS]

Homework Statement

A body is thrown into the air with an initial velocity of v₀. What initial velocity is required to double the maximum height previously attained?

Homework Equations

The Attempt at a Solution

I found the max height v₀²/64 at v₀ by solving for 't' in the velocity equation when 'v' = 0, and subbing that for 't' in the height equation and solving. I then multiplied this by 2 and set it to the height and tried to solve for v₀ and ended up with v₀² = -512t² + 32tv₀. I then solved for v₀ and ended up with v₀ = 16t. Do I need to somehow find 't' or did i stray off course somewhere and end up with the wrong answer?

A quick way to get the answer. (I'll put some subscripts on the variables.)

You found that throwing some body vertically into the air with a velocity, v₁,
will cause the body to attain a maximum height, h₁, (which is the height of the body, when the velocity becomes zero).

Your solution was: h₁ = v₁²/64, which is correct for a gravitational acceleration, g = 32 ft/s².

Since you solved this for arbitrary initial velocity, it's true that in general that a maximum height, h = v²/64 will be attained by a body thrown up vertically with velocity, v.

Solve this equation for the launch velocity, v, needed for the body to attain a height, h, above the ground.

v = √(64 · h) = 8·√(h).

Specifically, v₁ = 8·√(h₁).

So if you want h₂ = 2·h₁, then :

v₂ = 8·√(h₂)

= 8·√(2·h₁)​

To get a decent looking answer divide this equation by v₁ = 8·√(h₁).