Expert: Richard J. Raridon - 8/28/2009

Question
Hi!  I need help with a question about projectile motion found in an algebra packet.

"A projectile is launched straight up from ground level with an initial velocity of 320 ft/sec.

a. When will the projectile's height above ground be 1538 ft?

b. When will the projectile's height above ground be at most 1538 ft?

c. When will the projectile's height above ground be greater than or equal to 1538 ft?"

Is there an equation that goes along with this problem?  I understand that for part a, you must set the equation equal to 1538, for part b you must set the equation less than or equal to 1538, and for part c you must set the equation to be greater than or equal to 1538.

Thanks!
Jenny

Answer
The equation is H = v0t-(1/2)gt^2 where H is the height, v0 is the inital velocity and g is the acceleration due to gravity.  
So 1538 = 320t -(1/2)(32.2)t^2
rearrange to 16.1t^2 -320t +1538 = 0
Solve for t to get t = 8.14s and 11.74s
That's the answer for a.
b. t<8.14 and t>11.74
c. 8.14 < t < 11.74