Question

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?

Answer 1

t = 2.38 [s]

Step-by-step 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.

`x =x_(o) +v_(o) *t+((1)/(2) )*a*t^(2)`

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]