An object is dropped from a hot air balloon 500 meters above the ground. The height of the object after t seconds is given by s(t) = 500 − 4.9t2.

Question

What is the velocity (in meters per second) of the object when it strikes the ground? (Round your answer to two decimal places.)

Answer 1

s(t) = 500-4.9t^2

s'(t) = v(t) = -9.8t

when it hits the ground s(t) = 0 = 500-4.9t^2

49t^2 = 500

t^2 = 500/49

t = (10/7)sqr5 = about 3.1944 seconds to hit ground level

v(10sqr5/7) = -9.8(10sqr5)/7 m/s

= about -31.3 meters per second when it hits the ground

it's negative to indicate a downward velocity

the velocity is decreasing 9.8 m/sec/sec due to gravity

v= about -9.8 x 3.1944 = about -31.3 m/s