How Do You Calculate the Range of a Ball Thrown from a Hot Air Balloon?

[xXshortyXx]

Hey Guys
(sorry about other post in diff topic i never saw this one)
Here is the physics problem I am having problems with.

1)A .50 kg ball is thrown at 42 degrees above the horizontal at 19 m/s from a stationary hot air baloon 25m above the ground. What is the range?

ps Any general information on 2d kinomatics would be greatly appreciated. I am having trouble grasping the concepts in this chapter.

THX A MILLION!

 

[Chi Meson]

THis is a standard projectile motion problem. The "range" refers to the horizontal displacement, that is, the distance along the ground from a point directly underneath the balloon to the point where the ball lands.

We assume that air resistance is ignorable, so the mass of the ball is not important.

The time that the ball is in the air depends only on the vertical componants. So find the vertical component of the initial velocity. Since it goes up at first call this initial velocity positive.

The vertical displacement is the height of the balloon but since it ends up below the starting point, the vertical displacement is a negative value (it does not matter that the ball goes up higher before falling, the displacemnt only cares about the difference between the initial and final positions.

The vertical acceleration is -9.80 m/s^2 of course.

use the formula d=vt + 1/2 a t^2 and solve for t. This means you use the quadratic equation to find t. YOu get two answers, but only one will be positive. (Watch all the negative values, they are all important)

After you find the time of the flight, multiply this time by the horizontal component of the initial velocity. Since horizontal motion is not accelerated, d=vt. The d is your answer.

 

[Chi Meson]

Hey, Where's my million?

 

[xXshortyXx]

THX MAN that's exactly how its suppose to be done to you guys are great! Heres your million.

1 000 000

 

[Chi Meson]

Awesome