Question

A projectile is launched straight up at 135 m/s. Determine how fast it is moving at the top of its trajectory (flight).

Answer 1

It is 0 m/s. At the very top of its flight is right between when it is going up and going down; this is the ephemeral moment when it is perfectly still.

Answer 2

I do know that diagonal motion can be separated into separate horizontal and vertial components with trigonometry. You must use the information you are given as well as SOHCAHTOA, in cas you didn't know that means sine=opposite/hypotenuse, cosine=adjacent/hypotenuse, tangent=opposite over adjacent.

Answer 3

Basically, by making an imaginary right triangle and plugging your information into these equations, you can figure this stuff out.

Answer 4

The angle of the hypotenuse is angle of the object's velocity, and the length of the hypotenuse is the magnitude of the object's velocity. The adjacent side is EITHER 1)an imaginary line perpendicular to the direction of gravity if the object is going up; OR 2) an imaginary line parallel to the direction of gravity if the object is going down.

Answer 5

Since you know a right triangle has a right angle, you can figure out the opposite side after you have defined the hypotenuse and adjacent sides.

Answer 6

Physics problems usually assume you have access to a calculator that can automatically give you the sine, cosine, and tangent for any angle.

Answer 7

Well... thats it in a nutshell ... hope i helped

Answer 8

it seems that most of @woooooooooowwwww post is not truly relevant. However his FIRST post has it right.

You are told that it is fired vertically up - therefore it goes straight up, and straight back down again. At the very top it has no vertical velocity, and we know its horizontal velocity is zero too.