Projectile Motion -- position and the highest point

[Masrat_A]

For 3 (a), I feel like I've gotten it down correctly, but comments would be appreciated! 3 (b), I'm not too sure about; could anyone please give a thorough review and point out mistakes if there are any?

Number 4, I have no idea where to even begin. I've spent nearly an hour brainstorming, but have gotten nowhere. I do have class notes, but I'm unable to make sense of any of it. May I please have some direction on how I could go about solving? Are there any formulas I should be looking into?

1. Homework Statement
3) A projectile is fired from ground at 37 degrees above horizontal at speed 50 m/s.
a) Determine the position of the highest point and the time it takes to reach it.
b) Determine the position and time at which it strikes a wall 200 m away.

4) (See figure.) A lemming accelerates uniformly starting from rest 2 m before jumping horizontally from the top of a 20 m high cliff. What is the minimum acceleration required so that the lemming can clear the rocks stretching out 4 m from the base of the cliff?

Homework Equations

Figure for question 4: http://i.imgur.com/SMspyLdh.jpg

The Attempt at a Solution

http://i.imgur.com/MpjwWeih.jpg

 

[QuantumQuest]

I cannot make sense of some things in your solution. I'd recommend using Latex - there are few equations it is not difficult or tidy things up a little bit.

 

[MarkFL]

I tend to work these problems in general terms, so that we derive formulas that can be applied to similar problems in the future.

For problem 3, I would begin with:

$a_y=-g$

$a_x=0$

Resolving the initial velocity into $xy$-components, we have:

$v_{y_0}=v_0\sin(\theta)$

$v_{x_0}=v_0\cos(\theta)$

Now, using the fact that acceleration is the time rate of change of velocity, can you state the initial value problems (IVPs) whose solutions are the $xy$-components of the projectile's velocity and solve the IVPs?

 

[Tanishq Nandan]

Are there any formulas I should be looking into?

Equation of trajectory of a projectile in 2D.

For part 3(b),if you are familiar with the equation of trajectory of a projectile,that would very easily give you the position of the required point.

The time is pretty easy to calculate,since we know the horizontal component of the velocity(vcostheta) and distance travelled(200m)..then it's simply v=d/t

Question 4 can be broken into two parts..one projectile and one 1D kinematics
Once again,if you know what the equation of trajectory of a projectile is and how to apply it,the velocity will be quite easy to find.
Then comes the 1D kinematics part
Say the velocity you found out is v.
Your new question becomes "What constant acceleration should a lemming starting from rest have so that it can attain a velocity v within 2m??"
That should again be simple ,using Newton's equations of motion.

 

[Masrat_A]

I cannot make sense of some things in your solution. I'd recommend using Latex - there are few equations it is not difficult or tidy things up a little bit.

Here are my solutions to problem 3 (a) and (b) using latex.

$v_ox = 50\cos37^\circ = 40 = v_x$
$v_oy = 50\cos37^\circ = 30$
$x = 40t$

$v_y = 30 - 10t$
$y = 30t - t5^2$
$v_y^2 = 900 - 20y$

Solution for (a):

$v_y = 0$
$30 - 10t = 0$
$t = 3$ sec

$y = 20(3) - 5(3)^2 = 15 m$
$x = 40(3) = 120 m$
$(120, 15)$

Solution for (b):

$x = 40 t = 200$
$t = 5$ sec

$y = 20(5) - 5(5)^2 = -25$
$x = 40(5) = 200$
$(200, -25)$

Equation of trajectory of a projectile in 2D.

For part 3(b),if you are familiar with the equation of trajectory of a projectile,that would very easily give you the position of the required point.

The time is pretty easy to calculate,since we know the horizontal component of the velocity(vcostheta) and distance travelled(200m)..then it's simply v=d/t

Question 4 can be broken into two parts..one projectile and one 1D kinematics
Once again,if you know what the equation of trajectory of a projectile is and how to apply it,the velocity will be quite easy to find.
Then comes the 1D kinematics part
Say the velocity you found out is v.
Your new question becomes "What constant acceleration should a lemming starting from rest have so that it can attain a velocity v within 2m??"
That should again be simple ,using Newton's equations of motion.

At this moment, unfortunately, I am not very much familiar with equation of trajectory of a projectile. Is there any other method by which I could find the position for question 3b and velocity for question 4?

In addition, for 3b, would you please mind explaining which would be the horizontal component of velocity?

 

[SammyS]

That looks like the correct method, but you have a typo and are using an incorrect value as indicated below.

(You must be using 10 m/s² as a value for g.)

Here are my solutions to problem 3 (a) and (b) using latex.
$v_ox = 50\cos37^\circ = 40 = v_x$
$v_oy = 50\cos37^\circ = 30$
$x = 40t$

$v_y = 30 - 10t$
$y = 30t - t5^2$

You should have squared t, not g/2 .

$v_y^2 = 900 - 20y$

Solution for (a):

$v_y = 0$
$30 - 10t = 0$
$t = 3$ sec

$y = 20(3) - 5(3)^2 = 15 m$

You appear to have used a value of 20 m/s for $\ (v_0)_y \$ rather than 30 m/s.
You also did this for part (b).

$x = 40(3) = 120 m$
$(120, 15)$

Solution for (b):

$x = 40 t = 200$
$t = 5$ sec

$y = 20(5) - 5(5)^2 = -25$
$x = 40(5) = 200$
$(200, -25)$

For #4:
How much time does it take to fall 20 m ?

 

[Masrat_A]

That looks like the correct method, but you have a typo and are using an incorrect value as indicated below.

(You must be using 10 m/s² as a value for g.)
You should have squared t, not g/2 .You appear to have used a value of 20 m/s for $\ (v_0)_y \$ rather than 30 m/s.
You also did this for part (b).

Thank you for pointing those out! Here are my updated solutions; do these look better?

$v_ox = 50\cos37^\circ = 40 = v_x$
$v_oy = 50\cos37^\circ = 30$
$x = 40t$

$v_y = 30 - 10t$
$y = 30t - 5t^2$
$v_y^2 = 900 - 20y$

Solution for (a):

$v_y = 0$
$30 - 10t = 0$
$t = 3$ sec

$y = 30(3) - 5(3)^2 = 45 m$
$x = 40(3) = 120 m$
$(120, 45)$

Solution for (b):

$x = 40 t = 200$
$t = 5$ sec

$y = 30(5) - 5(5)^2 = 25$
$x = 40(5) = 200$
$(200, 25)$

For #4:
How much time does it take to fall 20 m ?

Assuming that the initial velocity is 0, this is what I have been able to come up with for time:

$d = V_i * t + \frac{1}2at^2$
$20 = \frac{1}29.8t^2$
$t = \sqrt{\frac{20}{4.9}}$
$t = 2.02$

Minimum acceleration:

$d = vt$
$4 = v * 2.02$
$v = \frac{4}{2.02}$
$v = 1.98 m/s$

$v_f^2 = v_i^2 + 2ad$
$v_f^2 = 3.92$
$3.92 = 2 * ax = 4 * a$
$a = \frac{3.92}{4}$
$a = 0.98 m/s^2$

 

[QuantumQuest]

$v_oy = 50\cos37^\circ = 30$

Be careful here $v_oy = 50\sin37^\circ$

Now, $y = v_0\sin(\theta) t - \frac{1}{2}gt^2$ as you wrote it (with values substituted) but as SammyS points out you must square $t$ not $\frac{g}{2}$ . We usually take $g = 9.8 \frac{m}{s^2}$ unless specified otherwise. Also, as mentioned, you must put $30 \frac{m}{s}$ in the equation of $y$ for the initial velocity component $v_0y$. Also, for part (b) the time you find is correct but again $v_0y$ must be corrected.

 

[SammyS]

Thank you for pointing those out! Here are my updated solutions; do these look better?

$v_ox = 50\cos37^\circ = 40 = v_x$
$v_oy = 50\cos37^\circ = 30$
$x = 40t$

$v_y = 30 - 10t$
$y = 30t - 5t^2$
$v_y^2 = 900 - 20y$

Solution for (a):

$v_y = 0$
$30 - 10t = 0$
$t = 3$ sec

$y = 30(3) - 5(3)^2 = 45 m$
$x = 40(3) = 120 m$
$(120, 45)$

Solution for (b):

$x = 40 t = 200$
$t = 5$ sec

$y = 30(5) - 5(5)^2 = 25$
$x = 40(5) = 200$
$(200, 25)$
Assuming that the initial velocity is 0, this is what I have been able to come up with for time:

$d = V_i * t + \frac{1}2at^2$
$20 = \frac{1}29.8t^2$
$t = \sqrt{\frac{20}{4.9}}$
$t = 2.02$

Minimum acceleration:

$d = vt$
$4 = v * 2.02$
$v = \frac{4}{2.02}$
$v = 1.98 m/s$

$v_f^2 = v_i^2 + 2ad$
$v_f^2 = 3.92$
$3.92 = 2 * ax = 4 * a$
$a = \frac{3.92}{4}$
$a = 0.98 m/s^2$

That looks good for both problems.

It's interesting that you have switched to using a value for g of 9.8 m/s² for problem 4.