Projectile Motion: only g and t known

[joshmdmd]

Homework Statement

There is a hot air balloon on the ground. It moves upwards at a constant acceleration. After 30 seconds someone throws a ball horizontally. An observer times the ball in the air to be 10 seconds.

Knowns:
Time of the ball: 10seconds
Gravity: 9.8m/s

You are to find:
a) the balloons acceleration
b) initial height at which the ball is thrown
c) height of the balloon when the ball hits the ground
d) ask God for forgiveness

Answers
a) 0.65m/s2
b) 294m
c) 522m

Homework Equations

vf^2=vi^2+2ad
d=vi(t)+0.5(a)t^2
d=vf(t)+0.5(a)t^2
d=vt
vf=vi+a(t)
d=0.5(vi+vf)t

The Attempt at a Solution

I tried to combine formulas but when I did i had gotten incorrect answers.
There are not enough knowns to solve it directly. The class (20 people) couldn't get it after 30min.. This is Grade 12 Physics..

I know that because the ball is moving horizontally then the velocity going horizontal is constant, and that because as it is thrown the balloon is moving upwards then the ball also has a velocity upwards, creating a parabola.

 

[joshmdmd]

any ideas would help really.. i have to figure this out and hand it in tommorow.

 

[phc]

Homework Statement

There is a hot air balloon on the ground. It moves upwards at a constant acceleration. After 30 seconds someone throws a ball horizontally. An observer times the ball in the air to be 10 seconds.

Knowns:
Time of the ball: 10seconds
Gravity: 9.8m/s

You are to find:
a) the balloons acceleration
b) initial height at which the ball is thrown
c) height of the balloon when the ball hits the ground
d) ask God for forgiveness

Answers
a) 0.65m/s2
b) 294m
c) 522m

Homework Equations

vf^2=vi^2+2ad
d=vi(t)+0.5(a)t^2
d=vf(t)+0.5(a)t^2
d=vt
vf=vi+a(t)
d=0.5(vi+vf)t

The Attempt at a Solution

I tried to combine formulas but when I did i had gotten incorrect answers.
There are not enough knowns to solve it directly. The class (20 people) couldn't get it after 30min.. This is Grade 12 Physics..

I know that because the ball is moving horizontally then the velocity going horizontal is constant, and that because as it is thrown the balloon is moving upwards then the ball also has a velocity upwards, creating a parabola.

Balloon:
y=(1/2)a.30^2; vy=30a;

Ball:
y=10vy - (1/2)g.10^2;

Thus: -(1/2)a.30^2=10vy-(1/2)g.100 = 10.30a - (1/2)g100
a = 0.654 m/s^2
etc.

 

[joshmdmd]

dam.. i figured it was something like that.. i figured it out the in the morning the next day and yah.. fun stuff =]