Air ballon and ball question. HELP

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Homework Help Overview

The problem involves a balloon ascending at a speed of 21.5 m/s while a ball is thrown upwards at 13.2 m/s from within the balloon. The ball is observed to hit the ground after 34.5 seconds. The questions posed relate to the time taken for the ball to reach its highest point, the height it achieves, and its speed upon impact with the ground.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to apply the principle of relative motion to determine the ball's initial velocity relative to the ground. There are questions about the correctness of the initial calculations and whether the equations used are appropriate.

Discussion Status

Some participants suggest that the velocities of the balloon and the ball should be combined to find the ball's velocity relative to the ground. Guidance is offered on how to approach the calculations for height and speed, while noting the need to consider the initial height of the balloon at the moment of release.

Contextual Notes

Participants highlight that the questions may be out of order and that the total time of 34.5 seconds should be used to backtrack and find the initial height from which the ball was released.

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Air balloon and ball question. HELP!

Homework Statement


A balloon is going straight up at 21.5 m/s. You are in the balloon and you throw a ball at 13.2 m/s straight up while riding in the balloon. If the ball hits the ground at 34.5s

a) How long oes it take the ball to stop once it leaves your hand at the ball's highest point?

b) How high did the ball go after leaving your hand? (from your hand to the highest point)

c) How high off the ground did the ball go?

d) How fast was the ball traveling when it hit the ground? (Not zero)

e) How high off the ground was the ball when it was released

Homework Equations


Vavg=(x/t)
x=1/2 (Vi+Vf)t
a= (Vf-Vi)/t
Vf=Vi + at
Vf^2= Vi^2 + 2ax
X= Vi t + 1/2(at)^2

a=-g=-9.81



The Attempt at a Solution


a) Vi= 13.2 m/s g=9.81 Vf= 0m/s t=?
t= (Vf-Vi)/a
t= (13.2-0)/9.81
t= 1.35 Seconds

b) x= 1/2 (Vi+Vf) t
x= 1/2 (13.2) 1.35
x= 8.88 m

c) Not sure yet

d) A guess
Vf= -9.81 (34.5) + 0
Vf= -338.445 m/s

e) no idea

Any help/corrections are appreciated to better my understanding. Thank you
 
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You need to use the principle of relative motion to get the correct initial velocity. The velocity of the ball relative to the balloon is 13.2 m/s, but the balloon is also has a velocity...
 


Is there an equation for that? Are my initial answers wrong as far as the equations used i just have the values wrong?
 


Yes, the velocity's are simply added together to get the velocity of the ball relative to the ground or reference frame. From that point all the steps should be the same. Set your velocity final to 0m/s to find the time it takes to reach a peak. Use that time to calculate the height it reached. The only thing is there is some initial height when the ball is released (question E) and you only calculated the height it goes up from that release point. So you need to use the total time given to back calculate that initial height. Then use that total height (or just the initial height plus the initial velocity) to calculate the speed it would reach before hitting the ground (question C). So it seems as the questions are out of order, but that's how I'd do it.
 

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