# Thread: Solving: Time for an object to hit the ground

1. ## Solving: Time for an object to hit the ground

So my exam is in 2 days and I am having so much trouble. I am a new member, thanks for taking the time to view this question.

If an object is thrown downward at 26 m/s from the top of a building, 102m high, the equation relating to displacement (meters) and time (seconds) is:

s = 1/2 gt^2 + 26t + 102

Find the time for an object to reach the ground from the top. Use g=-9.81

So I have come up with: (I am new to this; sq = square root symbol, sorry)

-26 +/- sq (26)^2 - 4(-4.905)(102) / -9.81

My problem occurs when my two answers, rougly, equal:

s= -7.9247
and
s= -2.624

I am confused as to why I have 2 negative times. One should be positive and one negative, correct?

Thank you

2. ## Re: Solving: Time for an object to hit the ground

Originally Posted by polskon
So my exam is in 2 days and I am having so much trouble. I am a new member, thanks for taking the time to view this question.

If an object is thrown downward at 26 m/s from the top of a building, 102m high, the equation relating to displacement (meters) and time (seconds) is:

s = 1/2 gt^2 + 26t + 102

Find the time for an object to reach the ground from the top. Use g=-9.81

So I have come up with: (I am new to this; sq = square root symbol, sorry)

-26 +/- sq (26)^2 - 4(-4.905)(102) / -9.81

My problem occurs when my two answers, rougly, equal:

s= -7.9247
and
s= -2.624

I am confused as to why I have 2 negative times. One should be positive and one negative, correct?

Thank you
The initial velocity is $-26 {\text{ m/s}$ (you are working with s positive upwards) and $g=-9.81 {\text{ m/s}}^2$, so your equation for displacement is:

$s(t)=- \frac{9.81t^2}{2}-26t+102$

which I can tell you has exactly one positive root (and in consequence one negative root, but that is not the one you want).

CB

3. ## Re: Solving: Time for an object to hit the ground

Thank you.

So am I supposed to adjust the formula based on what the question is asking? (A downward throw, for example)

4. ## Re: Solving: Time for an object to hit the ground

Originally Posted by polskon
Thank you.

So am I supposed to adjust the formula based on what the question is asking? (A downward throw, for example)
No you are supposed to use the values from the problem for the constants in the standard formula.

The displacement equation under constant acceleration is:

$s(t)=\frac{at^2}{2}+v_0t+s_0$

where $a$ is the acceleration, $v_0$ the initial velocity and $s_0$ the initial displacement.

In this problem $a=-9,81\ {\rm{m/s}}^2$, $v=-26\ {\rm{m/s}}$ $s_0=102\ {\rm{m}}$.

The minus signs are because we have chosen positive to indicate upwards and ground level to be $s=0$

CB