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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