# The flight time refers to the total amount of time the ball is in the air, from just after it is launched (t0) until just before it lands (t2). Hence the flight time can be calculated as t2âˆ’t0, or just t2 in this particular situation since t0=0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air.

**By:**Answerout

Here is the answer for the question - **The flight time refers to the total amount of time the ball is in the air, from just after it is launched (t0) until just before it lands (t2). Hence the flight time can be calculated as t2âˆ’t0, or just t2 in this particular situation since t0=0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air.**. You'll find the correct answer below

### The flight time refers to the total amount of time the ball is in the air, from just after it is launched (t0) until just before it lands (t2). Hence the flight time can be calculated as t2âˆ’t0, or just t2 in this particular situation since t0=0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height ymax, how long would it take to reach the ground? Ignore air resistance. Check all that apply. t0 t1âˆ’t0 t2 t2âˆ’t1 t2âˆ’t02

**The Correct Answer is**

**t1âˆ’t0t2âˆ’t1t2âˆ’t02(In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground.)**

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