Question #cd938

We're (essentially) asked to find the time when the rock hits the ground, given it fell from a height of #11.5# #"m"#.

This rock is undergoing free-fall; i.e. it is under the sole influence of Earth's gravitational force, and falling toward Earth's surface with an acceleration of #-g#, which is #-9.8"m/s"^2#

To find the time #t# when the rock reaches a position of #y = -11.5# #"m"#, we can use the equation

#y = y_0 + v_(0y)t - 1/2g t^2#

where

• #y# is the position at time #t# (#-11.5# #"m"#),

• #y_0# is the initial position (#0# #"m"#),

• #v_(0y)# is the initial #y#-velocity, which is #0# since the rock was merely dropped,

• #t# is the time, in #"s"# (what we must find), and

• #g# is #9.8# #"m/s"^2# (the minus sign in front of the #1/2# indicates that this acceleration is downward)

Plugging in known values, we have

#-11.5# #"m" = 0# #"m" + (0# #"m/s)"t - (4.9"m/s"^2)t^2#

#(4.9"m/s"^2)t^2 = 11.5# #"m"#

#t^2 = (11.5"m")/(4.9"m/s"^2)#

#t = sqrt((11.5cancel("m"))/(4.9cancel("m")"/s"^2)) = color(red)(1.53# #color(red)("s"#

When dropped from a well with a depth of #11.5# meters, the rock will thus take #1.53# seconds to reach the bottom of the well.