Hang-glider and water balloon problem

[Karajovic]

Homework Statement

A hang-glider, diving at an angle of 57.0^o with the vertical, drops a water balloon at an altitude of 680.0 m. The water balloon hits the ground 5.20 s after being released.
a. What was the velocity of the hang-glider?

Homework Equations

The Attempt at a Solution

we know that the initial speed of the water balloon is the speed of the hang-glider... so:
∆y = vi(∆t)sinx - 1/2g∆t^2
680 = vi(5.2)sin57 - 1/2(-9.81)(5.2)^2
vi = 186 m/s

but my textbook says the answer is 193 m/s.

Is the book wrong? or am I wrong?

thanks!

 

[tiny-tim]

Hi Karajovic!

A hang-glider, diving at an angle of 57.0^o with the vertical

so is it sin or cos?

(btw, are you ok on the other thread?)

 

[Karajovic]

Hey tiny-tim!

Thanks so much! I've been trying to find a solution for a while now! But may you please explain why you use cos?

Thanks again! (and I thought I replied to the other thread, but I guess it didn't send, I'll post there now!

 

[tiny-tim]

Hey Karajovic!

But may you please explain why you use cos?

You're probably used to being given the angle from the horizontal, in which case of course the horizontal component is cos, and the vertical component (which you need here) is sin.

The general rule is that it's always cos of the angle between the force and the direction.

So just ask yourself, is the given angle, θ, the angle between the force and the direction, or is it (90º - θ)?

Either way, it's cos … cosθ, or cos(90° - θ).

 

[Karajovic]

Hey Karajovic! You're probably used to being given the angle from the horizontal, in which case of course the horizontal component is cos, and the vertical component (which you need here) is sin.

The general rule is that it's always cos of the angle between the force and the direction.

So just ask yourself, is the given angle, θ, the angle between the force and the direction, or is it (90º - θ)?

Either way, it's cos … cosθ, or cos(90° - θ).

Argh... I don't get it... So, if you are given an angle with the horizontal, you use sin (but you said this is the vertical?) and if you are with the vertical, you use cos (which you said is the horizontal?)

What do you mean by the angle between the force and the direction?

(I put an attachment of a drawing... If angle B is 57, BC is 680 m and A is the direction of the hang-glider, which is the force? it can't be BC?)

Sorry if I may have created some confusion..

Thanks again!

 

[tiny-tim]

What do you mean by the angle between the force and the direction?

I mean if you're asked for the component of a force in a direction,

it's always cos of the angle between that force and that direction.

Sometimes they give you the wrong angle (usually 90° minus the angle) …

then it's cos of the correct angle, which is sin of the angle they gave you.

 

[Karajovic]

I mean if you're asked for the component of a force in a direction,

it's always cos of the angle between that force and that direction.

Sometimes they give you the wrong angle (usually 90° minus the angle) …

then it's cos of the correct angle, which is sin of the angle they gave you.

Oh! I apologize, yes, know I understand! You can still use sin, but since the angle is with the vertical you would subtract it by 90 to get your horizontal angle..

so sin(90-x) = cosx (x is with the vertical) ?

Thank you!