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

[piknless]

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.

 

[jaumzaum]

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

 

[Draco27]

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

 

[CAF123]

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 $H.$

 

[Nickie]

I've also tried using the given time and distance to solve for the acceleration (g) and then using that to solve for the initial height. However, both of these attempts have not yielded the correct answer.

Your attempts are on the right track, but there are a couple of things to consider in order to arrive at the correct answer. First, it's important to remember that the equation s=(1/2)(g)t^2 only applies when the initial velocity is zero and the object is dropped from rest. In this case, we are not given any information about the initial velocity, so we cannot use this equation directly.

Instead, we can think about the motion of the object in terms of its initial and final velocities. We know that the final velocity is zero when the object reaches the ground, and we can use the equation v = u + at (where v is final velocity, u is initial velocity, a is acceleration, and t is time) to find the initial velocity.

Since the final velocity is zero, we can rearrange the equation to solve for u:

u = -at

Now, we can use this value for u in the equation for displacement (s = ut + (1/2)at^2) to find the initial height:

s = ut + (1/2)at^2

x = (-at)(1.5) + (1/2)at^2

x = (-9.8)(1.5) + (1/2)(9.8)(1.5)^2

x = -14.7 + 11.025

x = -3.675 meters

Therefore, the total height the object was dropped from is 3.675 meters. It's important to note that this answer assumes that the acceleration due to gravity (g) is -9.8 meters per second squared, which is the standard value used in free fall calculations. If the problem specifies a different value for g, the answer will be different.