A ball dropped and accelerates downwards at a rate of 10m/s^2 for 15 seconds how much will the balls velocity increase

v = 150 m/s

Explanation:

Given that,

The acceleration of the ball, a = 10 m/s²

Time, t = 15 s

We need to find the increase in ball's velocity. When it is dropped its initial velocity is equal to 0. Let v be the velocity,

v = u+at

v = 0 +10(15)

v = 150 m/s

So, the ball's final velocity is 150 m/s.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west