Question

if you were to throw a large log over the edge of the grand canyon and took 5.65 seconds to hit the ground, calculate the velocity of the log at impact in m/s and calculate the distance the log fell in feet

Answer 1

Okay assuming the log had no initial vertical motion (you threw it straight out), the distance to the bottom is 

dist = 1/2 gt^2 = 1/2 x 9.8m/s/s x
(5.65s)^2 = 156.4m

the velocity at the bottom = g t = 9.m/s/s x 5.65s =55.4m/s 

Hope this helps! 

Answer 2

Answer:

55.4265 m/s

513.71 ft

Explanation:

t = Time taken to hit the ground = 5.65 seconds

a = Acceleration due to gravity = 9.81 m/s²

u = Initial velocity = 0 m/s

v = Final velocity

From equation of motion

$v=u+at\\\Rightarrow v=0+9.81\times 5.65\\\Rightarrow v=55.4265\ m/s$

∴ Velocity of the log at impact is 55.4265 m/s

Distance

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 5.65+\frac{1}{2}\times 9.81\times 5.65^2\\\Rightarrow s=156.5798625\ m$

Converting to feet

1 m = 3.28084 ft

156.5798625 m = 156.5798625×3.28084 ft

= 513.71 ft

∴ The distance the log fell is 513.71 ft