You throw a ball straight down off the roof of a 47 m tall building with an initial velocity of -8m/s. When does the ball hit the ground?

t = 2.38 [s]

Explanation:

To solve this problem we must use the following equation of kinematics. We must clarify that both the acceleration and the initial velocity were taken as positive, since the velocity of the movement coincides with the direction of the acceleration.

where:

x - Xo = distance = 47 [m]

Vo = initial velocity = 8 [m/s]

a = gravity acceleration = 9.81 [m/s²]

t = time [s]

Now replacing these values in the equation:

47 = 8*t + 0.5*9.81*t²

47 = 8t + 4.905t²

47 = 4.905*t(1.63 + t)

9.58 = t*(1.63 + t) solving this equation (cuadratic)

we found that t = 2.38 [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

there are no components of a physical goal mentioned on the list of choices that you've provided.