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for b...you can find the maximum height by taking the derivative, solving for the zeros of the derivative and doing the first derivative test to find the maximum
a) since the projectile can only be seen from above the rim of the gorge, \[-16t^2+175t-240\ge 240\] solve this and take only the positive values of t
^^ if any
u should find that the maximum is when t=175/32
I calculated a max height of 238.5, but does that mean it would always fall short of the 240 foot height rim of the gorge?
So there is no interval during which the projectile would be visible at ground level, right?
when you solve the equation given for part a greater than 240...it will never exist
b) the maximum height will be at -32t+175=0 that is t=175/32 therefor the maximum height is \[H_\max= v_i+s(175/32)\]
wait...i just realized that the equation is given from ground level
so the maximum height the is 238.51 above ground level
so part a is written wrong
you should be solving −16t2+175t−240≥0 for part a
I am making so many mistakes :(
for part a you should get t needs to be between 1.6078 and 9.3297
dont worry....i misread the problem at first too
But why isn't the max height then 478.5 (240+238.5), if its 238.5 feet above ground level. How can s(0) be at ground level when its being shot from 240 feet underground? I am so confused.
its not above ground until 1.6078
im guessing that part b is the max height from ground and not from the gorge since the equation given is from the ground