Projectile physics homework problem

[UNknown 2010]

Hello,

I have solved the following problem but my answer does not matched with the answer which is written in the book. My answer is 32 m/s but the answer of the book is 37 m/s

[PLAIN]http://img842.imageshack.us/img842/6706/projectile.png

Could anyone please show me where my mistake is ?


Thanks =)

 

[Delphi51]

I don't get either your answer or the book answer!
Maybe you will see my mistake . . .

I did not trust that v² formula because the signs are complicated: you would get the 32.08 answer regardless of whether the initial velocity was up or down! So for the vertical part I used d = vt+.5gt² with d = -50, v = +7sin(53), g= -9.81 to get that the time to fall is 3.37 s. Then I put that into
Vf = Vi + at = 7sin(53)-9.81*3.37 = -27.5 m/s.
Final answer of 27.8 m/s.

 

[collinsmark]

Hello,

I have solved the following problem but my answer does not matched with the answer which is written in the book. My answer is 32 m/s but the answer of the book is 37 m/s

[PLAIN]http://img842.imageshack.us/img842/6706/projectile.png

Could anyone please show me where my mistake is ?Thanks =)

I believe the answer you gave above (your answer) is correct, with possible, minor precision issues. The mistake must be in the book.

 

[Liquidxlax]

i got exactly your answer Unknown 2010, so the book is wrong

and Delphi51 you don't even need to look at time

 

[The legend]

^^same thing with me...the book seems to be wrong..

 

[collinsmark]

So for the vertical part I used d = vt+.5gt² with d = -50, v = +7sin(53), g= -9.81 to get that the time to fall is 3.37 s.

You might want to check your numbers again. I got t = 3.813 seconds using your method.

 

[Delphi51]

I still don't see it, Collinsmark. Hope you will pinpoint it for me!
-50 = 7*sin(53) - .5*9.81*t²
4.905*t² = 7*sin(53) + 50
4.905*t² = 55.59
t² = 55.59/4.905 = 11.33
t = 3.367

edit - Oops: forgot the t! Now getting the 31.8 m/s for the vertical.
Thank you for sorting that out!

 

[Liquidxlax]

I still don't see it, Collinsmark. Hope you will pinpoint it for me!
-50 = 7*sin(53) - .5*9.81*t²
4.905*t² = 7*sin(53) + 50
4.905*t² = 55.59
t² = 55.59/4.905 = 11.33
t = 3.367

edit - Oops: forgot the t! Now getting the 31.8 m/s for the vertical.
Thank you for sorting that out!

i honestly don't see how your solution is plausible of working. Can you explain how you take into account the time traveled above 50m?

 

[Delphi51]

The t is the time of flight. Those formulas Vf = Vi + at and
d = Vi*t + ½a*t² are for all constant acceleration motion and always account for the time, distance and speed automatically whether above or below the starting point. If you solve the first formula for t and sub into the second, you'll get the Vf² = Vi² + 2ad formula.

 

[zgozvrm]

Hello,

I have solved the following problem but my answer does not matched with the answer which is written in the book. My answer is 32 m/s but the answer of the book is 37 m/s

[PLAIN]http://img842.imageshack.us/img842/6706/projectile.png

Could anyone please show me where my mistake is ?


Thanks =)

Your work is perfect, with the minor exception that $$v_y$$ should be negative since the rock is moving downward at that point. This will also affect your angle. Remember that $$\sqrt{x^2} = \pm x$$

 

[Liquidxlax]

The t is the time of flight. Those formulas Vf = Vi + at and
d = Vi*t + ½a*t² are for all constant acceleration motion and always account for the time, distance and speed automatically whether above or below the starting point. If you solve the first formula for t and sub into the second, you'll get the Vf² = Vi² + 2ad formula.

okay i see what you did, and why it works