A stone is dropped out of a balloon which is descending at #14.7 ms^-1# when the balloon is at an altitude of #49 m#. How long before the stone hits the ground?

1 Answer
MattyMatty
Feb 5, 2018

#"2 seconds"#

Explanation:

#h = h_0 + v_0*t - g*t^2/2#

#h = 0 " (when stone hits ground, height is zero)"#
#h_0 = 49#
#v_0 = -14.7#
#g = 9.8#

#=> 0 = 49 - 14.7*t - 4.9*t^2#
#=> 4.9*t^2 + 14.7*t - 49 = 0#
#"This is a quadratic equation with discriminant : "#
#14.7^2 + 4*4.9*49 = 1176.49 = 34.3^2#
#=> t = (-14.7 pm 34.3)/9.8#
#"We have to take the solution with + sign as t > 0"#
#=> t = 19.6/9.8 = 2#

#h = " height in meter (m)"#
#h_0 = " initial height in meter (m)"#
#v_0 = " initial vertical velocity in m/s"#
#g = " gravity constant = 9.8 m/s²"#
#t = " time in seconds (s)"#

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