Vertical Motion Problem - Can't solve for (t): An object is dropped from a 112 foot building...

[threetimingalgebrafailure]

I've looked a lot online to find a way to solve the equation but couldn't anything. I hope anyone can help.
An object is dropped from a 112 foot building. When is the object 76 feet high?
g = 32 feet per second squared
v= 0 seconds
s =112 feet
h = 76 feet

I plugged in these values and get this equation:
h(t) = -1/2gt^2 + vt + s
76t = -1/2(32)t^2 + 0t + 112
76t = -16t^2 +112

I don't know how to solve for t. When I put it in the form of -16t^2 -76t + 112 = 0, I try factoring it to 4(-4t^2 -19 +28) = 0 but the solution is in radical form when the answer key says "The object is 76 feet high after 1.5 seconds."

Can anyone tell me what I'm doing wrong?

 

[Dr.Peterson]

I plugged in these values and get this equation:
h(t) = -1/2gt^2 + vt + s
76t = -1/2(32)t^2 + 0t + 112
76t = -16t^2 +112

You're misinterpreting function notation. When they write "h(t)", that doesn't mean "h times t"; it means "the value of h when the time is t", and is read "h of t" or "h at t".

So the equation should be 76 = -16t^2 +112.

Try taking it from there.

 

[stapel]

I've looked a lot online to find a way to solve the equation but couldn't anything.

To solve a quadratic equation, try using the Quadratic Formula.

An object is dropped from a 112 foot building. When is the object 76 feet high?
g = 32 feet per second squared
v= 0 seconds

Does "v" stand for "velocity"? If so, should the unit be "seconds", or perhaps "feet per second"?

s =112 feet
h = 76 feet

I plugged in these values and get this equation:
h(t) = -1/2gt^2 + vt + s
76t = -1/2(32)t^2 + 0t + 112

How are you getting that the height at the point they care about is 76t? Isn't the height given in terms of feet, not time?

76t = -16t^2 +112

I don't know how to solve for t. When I put it in the form of -16t^2 -76t + 112 = 0, I try factoring it to 4(-4t^2 -19 +28) = 0 but the solution is in radical form when the answer key says "The object is 76 feet high after 1.5 seconds."

Can anyone tell me what I'm doing wrong?

Try using the basic equation for projectile motion:

[imath]\qquad s(t) = -\frac{1}{2}gt^2 + v_0 t + h_0[/imath]

...where the variables are defined as follows:

[imath]\qquad s(t):\; \textrm{ height at time } t \textrm{ seconds}[/imath]

[imath]\qquad g:\; \textrm{ acceleration due to gravity}[/imath]

[imath]\qquad t:\; \textrm{ time, in seconds}[/imath]

[imath]\qquad v_0:\; \textrm{ initial velocity}[/imath]

[imath]\qquad h_0:\; \textrm{ initial height}[/imath]

The height is a function of time [imath]t[/imath], but [imath]s(t)[/imath] means that [imath]s[/imath] is a function of [imath]t[/imath], not that [imath]s[/imath] is multiplied by [imath]t[/imath].

 

[threetimingalgebrafailure]

You're misinterpreting function notation. When they write "h(t)", that doesn't mean "h times t"; it means "the value of h when the time is t", and is read "h of t" or "h at t".

So the equation should be 76 = -16t^2 +112.

Try taking it from there.

I got the answer. Thank you very much for helping me.

 

[Dr.Peterson]

I got the answer. Thank you very much for helping me.

Good. I figured that might be all you'd need.

 

[khansaheb]

I've looked a lot online to find a way to solve the equation but couldn't anything. I hope anyone can help.
An object is dropped from a 112 foot building. When is the object 76 feet high?
g = 32 feet per second squared
v= 0 seconds
s =112 feet
h = 76 feet

I plugged in these values and get this equation:
h(t) = -1/2gt^2 + vt + s
76t = -1/2(32)t^2 + 0t + 112
76t = -16t^2 +112

I don't know how to solve for t. When I put it in the form of -16t^2 -76t + 112 = 0, I try factoring it to 4(-4t^2 -19 +28) = 0 but the solution is in radical form when the answer key says "The object is 76 feet high after 1.5 seconds."

Can anyone tell me what I'm doing wrong?

If we place our origin of measurement at the top of the building, and measure 'y' downward positive,

The find of the problem is to calculate the time it will take to travel down the distance of (112-76=) 36 ft

Starting height = 0
starting time = 0

Initial velocity = 0
constant acceleration = 32 (positive downwards)

final distance = 36 = s
final time = t → find

equation to use → s = u*t + 1/2 * a * t^2

36 = 0*t + 1/2 * 32 *t^2 .....→..... t^2 = 36/16 .....→.... t = ?

 

[topsquark]

If we place our origin of measurement at the top of the building, and measure 'y' downward positive,

The find of the problem is to calculate the time it will take to travel down the distance of (112-76=) 36 ft

Starting height = 0
starting time = 0

Initial velocity = 0
constant acceleration = 32 (positive downwards)

final distance = 36 = s
final time = t → find

equation to use → s = u*t + 1/2 * a * t^2

36 = 0*t + 1/2 * 32 *t^2 .....→..... t^2 = 36/16 .....→.... t = ?

As an aside:
Whereas I constantly stress to Intro level students that we can pick any coordinate system we like, I usually suggest picking the lowest point on the sketch for the origin, and take the direction to the right and upward (away from the center of the Earth) as positive directions. I refer to it as the "usual coordinate system." My students have typically responded better to using this than any other coordinate system.

Then again, I usually also teach using the SI system!

-Dan

 

[topsquark]

And... [imath]g \neq 10 \text{ m/s}^2[/imath]. My students have to get calculators. They can use them!

-Dan

 

[Dr.Peterson]

If we place our origin of measurement at the top of the building, and measure 'y' downward positive,

I presume this question is from a math class, not physics, and they were just given this particular formula and this approximate value for g. In the absence of further information about background, I tend to go with what they are already doing, rather than suggest something very different.

Alternate methods, of course, are often worth offering after the immediate issue has been dealt with.

 

[khansaheb]

I presume this question is from a math class, not physics, and they were just given this particular formula and this approximate value for g. In the absence of further information about background, I tend to go with what they are already doing, rather than suggest something very different.

Alternate methods, of course, are often worth offering after the immediate issue has been dealt with.

I had thought all the immediate issues had been dealt with because the op posted:

I got the answer. Thank you very much for helping me.

Thus I proposed an alternate thought process.

 

[Dr.Peterson]

I had thought all the immediate issues had been dealt with because the op posted:


Thus I proposed an alternate thought process.

Yes, I almost ended with "... as they apparently have been", but for some reason I didn't. Maybe I decided that went without saying.

This was not intended as a criticism; I added the last sentence intending to clarify that, but I guess it didn't.