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anonymous
 5 years ago
GIVEN: A projectile is fired from the bottom of a 240 ft. deep gorge and is visible only when the projectile is above the rim of the gorge. If the projectile is fired with an initial velocity of 175 ft/s, the height of the projectile from ground level after t seconds is given by s(t) = 16t2 + 175t  240
(a) During what interval can the projectile be seen?
(b) What is the maximum height of the projectile?
anonymous
 5 years ago
GIVEN: A projectile is fired from the bottom of a 240 ft. deep gorge and is visible only when the projectile is above the rim of the gorge. If the projectile is fired with an initial velocity of 175 ft/s, the height of the projectile from ground level after t seconds is given by s(t) = 16t2 + 175t  240 (a) During what interval can the projectile be seen? (b) What is the maximum height of the projectile?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a) since the projectile can only be seen from above the rim of the gorge, \[16t^2+175t240\ge 240\] solve this and take only the positive values of t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u should find that the maximum is when t=175/32

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So there is no interval during which the projectile would be visible at ground level, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you solve the equation given for part a greater than 240...it will never exist

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b) the maximum height will be at 32t+175=0 that is t=175/32 therefor the maximum height is \[H_\max= v_i+s(175/32)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait...i just realized that the equation is given from ground level

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the maximum height the is 238.51 above ground level

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so part a is written wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you should be solving −16t2+175t−240≥0 for part a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[H_\max=V_i/t_o+s(t_0), t_0=175/32\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am making so many mistakes :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for part a you should get t needs to be between 1.6078 and 9.3297

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont worry....i misread the problem at first too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its not above ground until 1.6078

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im guessing that part b is the max height from ground and not from the gorge since the equation given is from the ground
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