An arrow is shot at a 30 degrees angle with the horizontal. It has a velocity of 49 m/s. How high up does it go?

First, determine how much of that 49m/s applies towards vertical velocity. So just multiply by the sine of 30. Giving

sin (30) * 49 m/s = 0.5 * 49 m/s = 24.5 m/s

Now the arrow will continue to climb until it's vertical velocity reaches 0. And since the attraction due to gravity is 9.8 m/s^2, that will be

24.5 m/s / 9.8 m/s^2 = 2.5 s

Finally, the distance an object travels under constant acceleration is given by the equation

d = 0.5 A T ^2

where

d = distance

A = acceleration

T = time.

So substitute the known values into the equation and solve for d.

d = 0.5 9.8 m/s^2 (2.5 s) ^2

d = 4.9 m/s^2 * 6.25 s^2

d = 30.625 m