How do you solve this problem? Bob shoots an arrow straight up with an initial velocity of 288 ft/sec from a h

Question

How do you solve this problem? Bob shoots an arrow straight up with an initial velocity of 288 ft/sec from a height of 45 feet. what is the arrow's maximum height and when does it reach that height? When will the arrow hit the ground?

Answer 1

(288^2)/(2*32.2) is the maximum height.
at time 288/32.2 seconds
it hits ground at time time t+576/32.2, where t is obtained from
45=288+.5*32.2*t^2

Answer 2

If I define g = 32 ft/s/s then the height reached is 45+ 288*288/(2*32) feet

it reaches that height 288/32 seconds after you fire it.

it hits the ground in just a bit more than twice that time cause it needs to go through the extra 45 feet but, neglecting air resistance, its doing 288 feet/sec on the way down as it goes through where you were standing. so the extra time is really tiny.

Answer 3

h = 45 + 288t - 16t² [orig. height + height from init velocity - pull of gravity]

h = 0 when the arrow is on the ground. Solve the quadratic:

45 + 288t - 16t² = 0
16t² - 288t = 45
16(t² - 18t ) = 45
16(t² - 18t + 81) = 45 + 16(81)
16(t - 9)² = 1341
(t - 9)² = 83.8125
t - 9 = ± 9.15
t = 9 ± 9.15
t = -0.15 s. or t = 18.15 s.

So the arrow hits the ground 18.15 seconds after Bob shoots it. Since the line of symmetry of the parabola is at t=9, that's where it reaches max height, ie,

h = 45 + 288(9) - 16(9)² = 1341 ft.

Answer 4

What the first guy said!!