A word problem for beg-int. physics....?

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Sweet_Lilly

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A bullet is fired upward from th ground at rest at a speed of 30m/s.
a hot air balloon is 12.0 m above the ground and traveing upward at a speed of 7.0 m/s.
identify two places where the altitude is the same for both objects. (1st on the way up and 2nd on the way down). what are their positions?
please answer or give pointers on how to get to the answer...not just the answer alone but in algebra/trig form (not calculus). explainations are appreciated so that i can understand how to do it in the future.
thanks!!!!!
 
Write out the positions of both objects as functions of time, then set the two expressions equal to each other and solve for the times that satisfy the equation.

I'll let my clock start at zero the instant the bullet is fired. I'll label ground level as zero height. Ignoring air resistance I can describe the bullet's height using the kinematic formula for constant acceleration --

y = y0 + v0*t - 1/2*g*t^2

where y0 = 0 because the bullet starts at ground level, v0 = 30 m/s, and g = 9.8 m/s^2. So the bullet's height is given by

y[bullet](t) = 0 + 30 m/s *t - 1/2*g*t^2

The balloon is simple rising at a constant rate of 7 m/s starting from 12 m, so the balloon's height is given by

y[balloon](t) = 12m + 7 m/s*t

To find the times when they are at the same height, insist that the two positions are equal for the same value of t:

0 + 30 m/s *t - 1/2*g*t^2 = 12m + 7 m/s*t

I use QuickMath to do my equation solving. Plugging in this equation (with the units taken out) gives

t = 0.597898 s

t= 4.09598 s
 
First working formula is

D1 = VoT + (1/2)gT^2

where

D1 = distance that bullet has travelled upward
Vo = initial velocity of bullet = 30 m/sec. (given)
T = time of travel
g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting values,

D1 = 30T - (1/2)(9.8)T^2

NOTE the negative sign attached to the acceleration due to gravity. This implies that the bullet is slowing down on its way up.

D1 = 30T - 4.9T^2 -- call this Equation 1

For the hot air balloon, an assumption will be made, i.e, it is travelling at a constant speed of 7 m/sec. Thus being said, its motion is not affected by the acceleration due to gravity.

Thus being said, for the hot air balloon,

D2 = 7(T) + 12 --- call this Equation 2

where

D2 = distance (from the ground) travelled by balloon when bullet passes it

Since Equation 1 = Equation 2, then

30T - 4.9T^2 = 7T + 12

4.9T^2 - 23T + 12 = 0

Using the quadratic formula,

T = 0.6 sec. and T = 4.1 sec.

The first location where the two objects will be the same distance from the ground will be

D2 = 7(0.6) + 12

D2 = 16.2 m

and the second location will be

D2 = 7(4.1) + 12

D2 = 40.7 m

Hope this helps.
 
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