Mistake in Physics textbook equation

[goochmawn314]

There is a formula in my textbook and I have a strong feeling it's incorrect.

The formula the book says is

velocity in y direction (final) = (vi)(sinx) - gt

Whenever I use this, I get all the answers wrong. However, if I use

(vi)(sinx) - 1/2(gt) .. I get them right.

Is this a misprint?

 

[Fightfish]

Probably not. It depends on how the angle is defined -- with respect to the horizontal, or with with respect to the vertical. These two definitions will affect whether the correct component is a sine or cosine of that particular angle.

 

[D H]

You haven't given the context, but I can guess: This is the expression for the vertical component of velocity of an object hitting the ground for an object thrown with an initial velocity v_i at an angle θ (measured with respect to horizontal).

If that's the case, your text is correct.

 

[td21]

I think the textbook is right for the following reasons:
1. vf = vi + at and in your case, a = -g in y direction
2. s = (vi)*t + 1/2*g*(t^2), but the LHS of your equation is velocity and the power of t in your equation is 1.

 

[goochmawn314]

Okay another equation in the book is

y = (vi)(sinx)(t) - 1/2(g)(t)^2 (unit here is meters)

If you divide out t on both sides you get meters over seconds and are left with (vi)(sinx) - 1/2gt

And meters/second is the unit for the original formula I posted. THe 1/2 is still there..

 

[goochmawn314]

also: the angle is always with respect to the horizontal in these problems unless it's aviation (with bearings)

 

[D H]

Okay another equation in the book is

y = (vi)(sinx)(t) - 1/2(g)(t)^2 (unit here is meters)

If you divide out t on both sides you get meters over seconds and are left with (vi)(sinx) - 1/2gt

And meters/second is the unit for the original formula I posted. THe 1/2 is still there..

The expression for altitude is correct. Your reasoning from that point on is incorrect. You need to take the derivative of the altitude to obtain the vertical component of velocity.

If you don't know what that means, you will eventually, but until then you'll have to take it for granted that the equations for altitude and the vertical component of velocity in your text are both correct.

 

[goochmawn314]

Yeah the derivative is just gt isn't it.. Ok but in this course, there hasn't been any calculus yet. Thanks for your help