Solve 1-D Motion Problems: Homework Statement

[Tchao]

Homework Statement

a. If a ball is launched straight upwards from the ground with an initial velocity of 20.0 m/s, how long does the ball take to reach a height of 15.0 m on the way up?

b. For the same situation, at what time does the ball reach 15.0 m on the way back down?

c. If a ball is dropped from a height of 100.0 m, what is it's height above the ground (in meters) after 3.0 s?

d. If a ball is dropped from a height of 100.0 m, what is it's velocity (in m/s) after 3.00 s? You may treat downward velocity as positive.

Homework Equations

Vx = V0x + axt
x= x0 +v0xt + 1/2axt^2

The Attempt at a Solution

a. 15 = 20t - 4.9t^2
I took the derivative of my equation.
15 = 20 - 9.8t
t = 0.510 s

b. 0 = 20t - 4.9t^2
I assume the height from problem a cancels out with problem b giving 0 and I also took the derivative of equation.
0 = 20 - 9.8t
t= 2.04 s

c. x = 100 - 4.9t^2
Plug in 3.0 s for t.
x = 55.9 m

d. Vx = 9.8 x 3.0 = 29.4 m/s

 

[kuruman]

Hi Tchao and welcome to PF.

I took the derivative of my equation.

Why? Just solve the quadratic and you will get two solutions. What is the meaning of each solution?

On edit: The derivative of 15 with respect to time is not 15.

 

[Tchao]

So, I get 3.09 s and 0.990 s when solving using the quadratic equation.
The 0.990 s would be the solution to problem A and the 3.09 s would be the solution to problem.

 

[kuruman]

So, I get 3.09 s and 0.990 s when solving using the quadratic equation.
The 0.990 s would be the solution to problem A and the 3.09 s would be the solution to problem.

Correct. The two solutions are the two times when the ball is at height 15 m. The last two parts look fine.

 

[PeroK]

The Attempt at a Solution

a. 15 = 20t - 4.9t^2
I took the derivative of my equation.
15 = 20 - 9.8t
t = 0.510 s

You cannot take the derivative here, because that is an equation for specific values of $t$.

For example, how long does it take to go $5m$ at $2m/s$?

You get the equation (1) $2t = 5$ hence $t = 2.5s$

But, if you differentiate that equation (1), you get the nonsensical $2 = 0$.

You can only differentiate something that is an equation for all $t$. For example: $s = ut + 0.5 at^2$ is an equation that holds for all $t$, where $u, a$ are constants and $s$ is, therefore, a function of $t$.

Differentiating that equation gives $v = \frac{ds}{dt} = u + at$, which is then valid for all $t$ and is, as you may recognise, another equation of motion.

 

[hmmm27]

The only thing left to say is puhlease use BBcode and/or LaTex, both of which are built right into the editor, are very easy to use, and the links to the guides are right under the input box, on the left.

 

[kuruman]

The only thing left to say is puhlease use BBcode and/or LaTex, both of which are built right into the editor, are very easy to use, and the links to the guides are right under the input box, on the left.

Being a first time user, OP may not be familiar with these niceties.

 

[hmmm27]

Being a first time user, OP may not be familiar with these niceties.

True, just pointing them out : rather new myself - it's great watching what would normally be indecipherable ascii scribbles turn into textbook-ready formulas.

 

[hmmm27]

For this example

x= x0 +v0xt + 1/2axt^2
turns into
$$d_t=d_0+v_0t+\frac{1}{2}at^2$$