at time t seconds after the bike passes, the distance z is
z^2 = (88t)^2 + (110+15t)^2 = 7969t^2 + 3300t + 12100
2z dz/dt = 15938t + 3300
at t=5, z = 477.31
so now you have
2*477.31 dz/dt = 82990
dz/dt = 86.94 ft/s
A hot-air balloon is 110 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 60 mi/hr (88 ft/s). If the balloon rises vertically at a rate of 15 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 5 seconds later?
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