Question

an arrow is shot at a 30 degree angle with the horizontal. It has a velocity of 49 m/s. how high does the arrow go?

Answer 1

According to eqaution \[\max height= \frac{ v ^{2}\sin ^{2} a }{ 2g }\]
where 'v' is velocity 'a' is the angle with horizontal and 'g' is acceleration due to gravity

so by substituting and calculating values we get max height=30.625 m

Answer 2

In order to help you, what don't you understand exactly?

Answer 3

I don't understand, either. Please , explain

Answer 4

Nice name @Loser66

Answer 5

hehehe, my name is a perfect excuse for "I don't understand" , right?

Answer 6

It depends on how you took it, its either memorizing equations and solve (like the one above that @maqson answered) or you actually understand what's going on and solve. I actually like the second way, because you can solve any question you want. The first way, is of course, faster and maybe easier.

Answer 7

then?

Answer 8

I'm assuming that's Uy?

Answer 9

ok

Answer 10

using this equation,
\[y=\frac{ 1 }{ 2 }a t^{2}+uyt\]
just substitute and you'll find y which is the height :)

Answer 11

You need to calculate t to use the expression above.

The vertical component of the velocity, (49 m/s) sin(30) will decrease by 9,8 m/s per sec.
Find how long it takes to get to zero and use that time in the distance formula.

Answer 12

My way is a bit complicated so just do what maqson did and you're on the safe side :)