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?

Question

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 · Answer

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

anonymous · Answer

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

anonymous · Answer

^^
if any

anonymous · Answer

u should find that the maximum is when t=175/32

anonymous · Answer

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?

anonymous · Answer

yes

anonymous · Answer

So there is no interval during which the projectile would be visible at ground level, right?

anonymous · Answer

when you solve the equation given for part a greater than 240...it will never exist

anonymous · Answer

b) the maximum height will be at -32t+175=0 that is t=175/32
therefor the maximum height is
$$H_\max= v_i+s(175/32)$$

anonymous · Answer

sorry

anonymous · Answer

wait...i just realized that the equation is given from ground level

anonymous · Answer

so the maximum height the is 238.51 above ground level

anonymous · Answer

so part a is written wrong

anonymous · Answer

−16t2+175t−240≥0

anonymous · Answer

you should be solving −16t2+175t−240≥0 for part a

anonymous · Answer

$$H_\max=V_i/t_o+s(t_0), t_0=175/32$$

anonymous · Answer

$$V_i*t_0$$*

anonymous · Answer

I am making so many mistakes :(

anonymous · Answer

for part a you should get t needs to be between 1.6078 and 9.3297

anonymous · Answer

dont worry....i misread the problem at first too

anonymous · Answer

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.

anonymous · Answer

its not above ground until 1.6078

anonymous · Answer

im guessing that part b is the max height from ground and not from the gorge since the equation given is from the ground