Freefall : speed and velocity?

Question

a ball was dropped from a very high building

a. what is speed of the ball and how far it is below its dropping point 5.0 seconds after it was dropped

b. how many seconds since the ball was dropped will it be 3.136 x 10exp4 cm below its dropping point? what will be its velocity at this

Answer 1

Disregarding friction

a.

(9.8 m/s^s) * (5.0 s)

=

49 m/s

b.

(313.6 m) / (4.9 m/s^2) = t^2

[(313.6 m) / (4.9 m/s^2)] ^ (1/2) = t

t = 8 s

(8 s) * (9.8 m/s^2)

=

78.4 m/s

Answer 2

Take the direction of gravity as positive direction

s = (1/2)gt^2 = 4.9 t^2

v = gt = 9.8 t

(a)

t = 5 s

s = 4.9 (0.5)^2 = 1.225 m

v = 9.8 (0.5) = 4.9 m/s

(b)

s = 3.136 x 10^4 cm = 313.6 m

313.6 = 4.9 t^2

t = sqrt (313.6 / 4.9) = 8 s

v = 9.8 x 8 = 78.4 m/s