How High Was the Object Dropped in This Free Fall Problem?

AI Thread Summary
An object falls the last 30 meters of a building in 1.5 seconds, prompting a calculation of the total height from which it was dropped. The problem requires using kinematic equations, recognizing that the initial velocity during the last segment is not zero. To solve, the motion is divided into two parts: the distance fallen before reaching 30 meters and the fall of the last 30 meters. By calculating the velocity at 30 meters and using it to find the time taken for the entire fall, the total height can be determined. This approach leads to the conclusion that the total height is approximately 37.81 meters.
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Homework Statement



A certain object is falling in free fall down a building. The object falls the last 30 meters in 1.5 seconds. What was the total height it was dropped at.

Homework Equations



s=(1/2)(g)t^2

The Attempt at a Solution



I've tried setting the total height to x+30 and solving for x.
 
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piknless said:

Homework Statement



A certain object is falling in free fall down a building. The object falls the last 30 meters in 1.5 seconds. What was the total height it was dropped at.

Homework Equations



s=(1/2)(g)t^2

The Attempt at a Solution



I've tried setting the total height to x+30 and solving for x.


Actually you can use (1/2)gt² only if the initial velocity is 0. In the last 30m the initial velocity is not 0.

You have 30 = V.t + gt²/2 where V is the velocity in h=30m.

You find V=12.5
So T = 12.5/10 + 1.5 = 2.75s
Now you use gt²/2 and you get 37.81m
 
lemme try

suppose total height is h+30

now let time be t(to cover h distance)
h=1/2 gt2
h+30=1/2g(t+1.5)2(it takes 1.5 sec more)

solve for h and t u should find both
 
piknless said:

Homework Statement



A certain object is falling in free fall down a building. The object falls the last 30 meters in 1.5 seconds. What was the total height it was dropped at.

Homework Equations



s=(1/2)(g)t^2

The Attempt at a Solution



I've tried setting the total height to x+30 and solving for x.

It is best to split this problem into two separate parts. Consider the motion between the top of the building until the ball is at a height of 30m above the ground (first part). Then consider the motion from this 30m height to the ground (second part).

If the total height fallen is H then 30m above the ground, it has traveled (H -30)m. It was dropped with zero initial velocity so for the first part of the motion, we have -(H-30) = -\frac{1}{2}gt^2,\,\, (1) keeping consistent with signs.
Now consider the second part. We want to determine the velocity at a height 30m above the ground. You know the time taken to fall this distance so using one of the kinematic relations you should be able to find this velocity.

Now we have all the information required to find the time taken to travel (H-30)m. This is found using another (different) kinematic relation. Sub this time into eqn (1) and you have your H.
 
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