Question

An arrow is shot vertically upward from a platform 44 ft high at a rate of 220 ft/sec. When will the the arrow hit the ground?

Answer 1

The arrow will hit the ground approximately 2.648 seconds after it is shot.

Step-by-step explanation:

To solve this problem, we need to use the equation of motion for an object in free fall. The equation is h(t) = h0 + v0t - 0.5gt^2, where h(t) is the height at time t, h0 is the initial height, v0 is the initial velocity, g is the acceleration due to gravity, and t is the time. In this case, h0 = 44 ft, v0 = 220 ft/sec, and g = 32 ft/sec^2.

Let's set h(t) = 0 and solve for t:

0 = 44 + 220t - 0.5(32)t^2

0 = 32t^2 - 220t - 44

Solving this quadratic equation gives us two solutions: t = 0.8317 sec and t = 2.648 sec.

Since the arrow is shot vertically upwards, we can ignore the negative solution. Therefore, the arrow will hit the ground approximately 2.648 seconds after it is shot.