Question

A ball is thrown directly downward with an initial speed of 7.95 m/s from a height of 30.6 m. after what time interval does it strike the ground?

Answer 1

1.82 seconds

Step-by-step explanation:

∆ y = 1/2at^2 + Vi*t
a is acceleration, y is height, and Vi is initial velocity
30.6 = 1/2*9.8*t^2 + 7.95 * t
simplify and rearrange
4.9t^2 + 7.95t - 30.6 = 0
solve the quadratic
t = 1.816128... and a negative number, but since this is physics, in this case time can't be negative
Round to 3 significant figures
t = 1.82