Hang-glider and water balloon problem

• Karajovic
In summary, the hang-glider dropped a water balloon at an altitude of 680.0 meters and the balloon hit the ground 5.20 seconds after being released. The hang-glider's initial speed was 189.2 meters per second. The book says the answer is 193.0 meters per second, but the hang-glider's velocity was actually 186.0 meters per second.
Karajovic

Homework Statement

A hang-glider, diving at an angle of 57.0o 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?

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!

Hi Karajovic!
Karajovic said:
A hang-glider, diving at an angle of 57.0o with the vertical

so is it sin or cos?

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

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!

Hey Karajovic!
Karajovic said:
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).

tiny-tim said:
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!

Attachments

• Untitled.png
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Karajovic said:
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.

tiny-tim said:
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!

1. What is the "Hang-glider and water balloon problem"?

The "Hang-glider and water balloon problem" is a thought experiment that looks at the relationship between air resistance and gravity by considering a person hanging from a hang-glider and throwing a water balloon at different angles while in flight.

2. What is the purpose of this thought experiment?

The purpose of the "Hang-glider and water balloon problem" is to illustrate the principles of air resistance and gravity, and how they interact with each other in a real-life scenario.

3. How does air resistance affect the trajectory of the water balloon?

Air resistance, also known as drag, acts in the opposite direction of motion and increases as the speed of the object increases. In the case of the water balloon, air resistance will cause it to slow down and change its trajectory, making it fall at a steeper angle than it was thrown.

4. What role does gravity play in this thought experiment?

Gravity is the force that pulls objects towards the center of the Earth. In this thought experiment, gravity is the force that pulls both the hang-glider and the water balloon towards the ground. It is also responsible for the acceleration of the objects as they fall.

5. How does the shape and weight of the water balloon affect the results of this experiment?

The shape and weight of the water balloon can significantly impact the results of this experiment. A heavier and more aerodynamic balloon will experience less air resistance and fall at a slower rate compared to a lighter and less aerodynamic balloon. The shape of the balloon can also affect its trajectory and how it interacts with air resistance.

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