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## Homework Statement

Hello everyone, first time poster, long time lurker here. I'm currently enrolled in an online grade 11 physics course and don't have immediate access to a teacher, nor do they seem as helpful and friendly as PF! My algebra skills are not the best, so I may need a little clarification during this process.

Anyways, the problem presented below has me a little confused, because in my book, I'm supposed to let displacement equal 0 and find time, but there are two instances of time in the equation. What I'm confused about is that the book shows that the first time ( [itex] \Delta t [/itex] ) just vanishes and doesn't explain why.

I'm assuming it's being brought to the other side of the equation by dividing by 0. There comes another problem, at least in my book where the second time ( [itex] \Delta t^2 [/itex] ) is being factored out, which I don't understand because why wouldn't you just isolate it and then bring the square over, which gives the answer 5.06 s, whereas, in the book it says the answer to be 0 and 6.2 s?

Am I completely wrong? Is this a typo in my book? Am I insane?

**A golf ball is hit with an initial velocity of 40 m/s at an angle of 50 degrees.**

a.) How long is the ball in the air?

a.) How long is the ball in the air?

## Homework Equations

[itex]\vec{v}_iv=\vec{v}_i sin \theta [/itex]

[itex]\Delta \vec{d}=\vec{v}_iv \Delta t +\frac {1}{2} \vec{a} \Delta t^2 [/itex]

## The Attempt at a Solution

[itex]\vec{v}_iv=(40 m/s) sin 50° [/itex]

[itex] \vec{v}_iv=30.6 m/s [up] [/itex]

[itex] 0=(30.6 m/s) \Delta t - 4.9 m/s^2 \Delta t^2 [/itex]

[itex] \Delta t= √ 25.7 [/itex]

[itex] \Delta t= 5.06 s [/itex]