In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:

a = -32 ft/s/s

v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)

Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)

t = ?? sec.

The equation we will use is the one for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

$-15=(0)t+\frac{1}{2}(-32)t^2$ which simplifies down to

$-15=-16t^2$ so

$t=\sqrt{\frac{-15}{-16} }$ so

t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)

Now we use that time in the x-dimension. Here's what we know in that dimension specifically:

a = 0 (acceleration in this dimension is always 0)

v₀ = 80 ft/sec

t = .968 sec

Δx = ?? feet

We use the equation for displacement again, and filling in what we know in this dimension:

Δx = $(80)(.968) +(0)(.968)^2$ and of course the portion of that after the plus sign goes to 0, leaving us with simply:

Δx = (80)(.968)

Δx = 77.46 feet

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3 years ago