Projectile motion help please. Exam tomorrow!

[esig]

I have an exam tomorrow and I have spent 3 hours trying to figure this out. I have the answer but not the method.

 

You throw a rock from an edge at the speed of 12 m/s at an angle of 25°.

You are 1,5 m above the edge and it is 10 m above the ground. No air resistance.

 

Have calculated max height and time 0,61 s and 3,3 m

 

b) how long is the rock in the air?

c) how far from the edge (horizontally) does the rock land?

d) what speed does the rock have when it lands?

 

 

Answers

b) 2,3 s (2,262)

c) 24 m (23,51)

d) 19 m (19,22)

 

Please help me, am a bit desperate, really want to pass this test!

I have figured out b and c

Still need help with d

[studiot]

presumably d is given in m/s since it is a velocity.

 

Have you drawn a diagram aand calculated the time of flight?

 

Velocity at time t has components

 

Vcos(a) and [Vsin(a) -gt]

 

Hence total velocity at time t is

 

[math]\sqrt {{V^2}{{\cos }^2}a + {{(V\sin a - gt)}^2}} [/math]

which can be simplified.

Edited January 4, 2015 by studiot

[imatfaal]

first thing is to draw a diagram

 

second make sure you have defined your questions - is max height above ground, above starting point,or above cliff edge? etc

 

thirdly figure out where you can divide your problem into more manageable portions (ie where is vertical velocity zero?)

 

fourthly - write out the equations you should need and the variables you have and are searching for

 

===

 

I do not agree with your first answers. Maybe I am too tired - but I get max height is 1.31m above launch (or 2.81 above cliff or 12.81 above floor) and time of max height as .517s

 

I use v=u+at v=0 u = vertical component of 12m/s = 12sin(25) a=-9.8m/s^2 t=?

 

and then I sub back into s=ut+1/2at^2

 

===

 

That gets you to the high point. you then have a simple trajectory question that I would do from there

[esig]

I do not agree with your first answers. Maybe I am too tired - but I get max height is 1.31m above launch (or 2.81 above cliff or 12.81 above floor) and time of max height as .517s

 

I use v=u+at v=0 u = vertical component of 12m/s = 12sin(25) a=-9.8m/s^2 t=?

 

and then I sub back into s=ut+1/2at^2

 

===

 

That gets you to the high point. you then have a simple trajectory question that I would do from there

Those are the answers given by my teachers.

 

 

But I have calculates time of flight, although I do not get the exact same number as given by my teachers but they are close. For instance I get 2,2449 seconds the rock is in the air but the teacher gave 2,262 s

And I get the distance as 23,3297 m, but I do get the same as the teacher if I use the same time he got witch is 23,51 m.

 

 

Max height is above starting point. I am not sure how to figure out where the vertical velocity is zero.

 

 

[math]\sqrt {{V^2}{{\cos }^2}a + {{(V\sin a - gt)}^2}} [/math]

 

which can be simplified.

 

How is it simplified?

Have figured it out. Thanks for your help

[studiot]

Simplified?

[math]\sqrt {{V^2} - 2gtV\sin a + {g^2}{t^2}} [/math]
Good luck in your exam.

[studiot]

Hi, I hope your exam went well.

 

Now that it is over and since there seems to be some question as to the numerical answers, I can post my full solution.

 

Note that the actual numbers will depend upon the value taken for g. I have used g = 9.8 since it make no difference to a calculator.

For hand calculation it is often a good dodge to use g =10.

 

As I recommended I have drawn a diagram, showing all the information given.

 

I also recommend writing down all the equations of motion and assembling a list of the known and unknown data.

 

You can then look at the list and see which equation has only one unknown to use.

 

I have done this in the first attachment.

 

and followed my own advice in the second.

 

I wonder, perhaps if you miscopied the original height and time of flight.

Note also the correct use of negative quantities.