Vertical and horizontal components of its velocity? projectiles?

Question

Here is the picture

http://farm4.static.flickr.com/3065/2816448096_1c9d74deeb_o.jpg

At the instant just before the projectile hits the ground,

what is the Vertical and horizontal components of its velocity? (51.9 m/s and -63.1 m/s) == How do u get this answer?

What is the magnitude of velocity? (81.7 m/s) == what is a magnitude and how to u calculate it?

What is the angle made by the velocity vector with the horizontal? (50.6 degree below horizon) == What angle does this mean or where is this angle?

i just want to know how to get the answer like what equations are we suppose to use for each question

Answer 1

First, calculate the initial components of velocity:

Vxo = 65 * cos (37) = 51.9 m/s

Vyo = 65 * sin (37) = 39.1 m/s

If we assume no air resistance the x-component of velocity is constant, so it's the same right before impact as Vxo (51.9 m/s)

For the y-component, first find the time of flight:

X = Vx * t; t = X / Vx

t = 541 / 51.9 = 10.4 sec

Then use the time of flight to find Vy, just use

Vy = Vyo - g * t

Vy = +39.1 - 9.8 * 10.4 = -62.8 m/s (down) (different from your given answer just due to rounding...)

The magnitude of this velocity is just

V^2 = 51.9^2 + (-62.8)^2

V = 81.5 m/s

When it hits the ground, the projectile is moving to the right at 51.9 m/s and down at 62.8 m/s.  The angle these two components make with the horizontal (ground) is just

TAN (angle) = Vy / Vx

TAN (angle) = 62.8 / 51.9

angle = 50.4 deg

Good Luck!