Asked by Nilan
height of a projectile thrown straight up from top of 480 ft hill t seconds after bing thrown up with initial velocity vo ft/s is
h = -16t^2 +vot + 480
If vo = 96, find times when the projectile will be at a) height of 592 ft above the ground and b) crash on the ground.
h = -16t^2 +vot + 480
If vo = 96, find times when the projectile will be at a) height of 592 ft above the ground and b) crash on the ground.
Answers
Answered by
Reiny
so h = -16t^2 + 96t + 480
a)
you want
-16t^2 + 96t + 480 = 592
t^2 - 6t + 7 = 0
(t-7)(t+1) =0
t = 7 or a negative
b)
same thing except now you want
-16t^2 + 96t + 480 = 0
solve for t, use only the positive result
a)
you want
-16t^2 + 96t + 480 = 592
t^2 - 6t + 7 = 0
(t-7)(t+1) =0
t = 7 or a negative
b)
same thing except now you want
-16t^2 + 96t + 480 = 0
solve for t, use only the positive result
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