SOLUTION: Height of a Projectile A projectile is fired straight upward from ground level with an initial velocity of 128 feet per second. During what time period will its height be less tha

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Question 180636: Height of a Projectile
A projectile is fired straight upward from ground level with an initial velocity of 128 feet per second. During what time period will its height be less that 128 feet?
Please help me with this. Thanks!

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A projectile is fired straight upward from ground level with an initial velocity of 128 feet per second. During what time period will its height be less that 128 feet?
;
The equation we use for this:
h = -16t^2 + 128t
where h is the height (in feet) in t seconds, launched at 128 ft/sec
:
Find the t when the projectile reaches 128 ft
-16t^2 + 128t = 128
;
-16t^2 + 128t - 128 = 0
Simplify and change the signs, divide equation by -16
t^2 - 8t + 8 = 0
:
Solve this using the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=1, b=-8, c=8 (x = t)
t+=+%28-%28-8%29+%2B-+sqrt%28-8%5E2+-+4+%2A1%2A8+%29%29%2F%282%2A1%29+
t+=+%288+%2B-+sqrt%2864+-+32+%29%29%2F%282%29+
t+=+%288+%2B-+sqrt%2832+%29%29%2F%282%29+
Two solutions:
t+=+%288+-+5.654685%29%2F%282%29+
t+=+2.34315%2F2
t = 1.17 sec, passes 128 ft on the way up
and
t+=+%288+%2B+5.654685%29%2F%282%29+
t+=+13.654685%2F2
t = 6.828 sec, passes 128 ft on the way down
;
Graphically it can be seen
+graph%28+300%2C+200%2C+-4%2C+10%2C+-50%2C+275%2C+-16x%5E2%2B128x%29+
Find when it strikes the ground
-16t^2 + 128t = 0
Factor
-16t(t - 8) = 0
t = 0
and
t = 8 sec
;
we can say it's less than 128 ft:
1.17sec - 0 = 1.17 sec
and
8 - 6.83 = 1.17 sec
for a total of 3.34 seconds
:
Hope this makes sense to you. studying the graph will help.