Thomas threw an apple to Mikey, who was on a balcony 40 ft above him. He tossed it upward with an initial speed of 56 ft/s. Mikey missed it on the way up but then caught it on the way down. How long was the apple in the air?

Type your answer to complete this sentence. If your answer is not a whole number, use a decimal.

The apple was in the air for [Fill in the blank],
seconds.

Question

Thomas threw an apple to Mikey, who was on a balcony 40 ft above him.  He tossed it upward with an initial speed of 56 ft/s.  Mikey missed it on the way up but then caught it on the way down.  How long was the apple in the air?

Type your answer to complete this sentence.  If your answer is not a whole number, use a decimal.

The apple was in the air for [Fill in the blank],
 seconds.

BlackTom AI · Accepted Answer

The problem describes a vertical motion scenario: an object is thrown upward with an initial speed of 56 ft/s from a point on the ground, and Mikey is 40 ft above that point. We model the vertical position y(t) (in feet) relative to the thrower with downward taken as negative and using g = 32 ft/s^2: y(t) = v0 t − (1/2) g t^2 = 56 t − 16 t^2.

To find when the apple reaches Mikey’s balcony, we set y(t) equal to 40 and solve for t:
56 t − 16 t^2 = 40
−16 t^2 + 56 t − 40 = 0
Multiply by −1 to simplify: 16 t^2 − 56 t + 40 = 0
Divide every term by 8: 2 t^2 − 7 t + 5 = 0
Use the quadratic formula: t = [7 ± sqrt(49 − 40)] / 4 = [7 ± sqrt(9)] / 4 = (7 ± 3) / 4.
This gives two times: t = (7 − 3)/4 = 4/4 = 1 second (on the way up) and t = (7 + 3)/4 = 10/4 = 2.5 seconds (on the way down).

Since Mikey catches the apple on the way down, the relevant time is the later root, 2.5 seconds.

Now, consider the provided answer option: 3.7.
- Why this option is incorrect: the exact solution to the equations above yields the catch time on the way down as 2.5 seconds, not 3.7. The calculation explicitly uses the standard kinematic equation for vertical motion with initial upward speed 56 ft/s and gravity 32 ft/s^2, equating the height to 40 ft and solving the resulting quadratic. The roots of the equation correspond to the two moments the apple is at 40 ft height; the later root must be taken for the catch on the way down.

In short, the derived times when the apple is at 40 ft are 1 s and 2.5 s; 3.7 s does not satisfy the motion equation with the given initial speed and height, so it cannot be correct for the moment of catch on the way down.

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