Tampa is bracing for a near-certain washout this Fourth of July with a 95% rain chance and 11.1mm of expected precipitation, the highest rain probability across the state today, according to National Weather Service data. Jacksonville, meanwhile, bakes at a holiday high of 95°F with a 0% rain chance, creating a 20-degree Fahrenheit spread between the two cities’ rain-soaked and sun-scorched Independence Day experiences.
In Tampa, winds are gusting to 13 mph alongside a 91°F high and a low of 78°F, with humidity at 69%. Orlando reaches 89°F with a 37% rain chance and 13.6mm of precipitation despite winds of only 9 mph. Jacksonville’s 95°F high contrasts with a low of 75°F and 59% humidity, while Miami sits at 87°F with a 33% rain chance, 67% humidity, and winds up to 15 mph.
Tampa’s rain chance drops sharply to 35% on Sunday July 05 before spiking back to 77% by Tuesday July 07, with a high of 91°F. Jacksonville climbs to a dangerous 99°F on Monday July 06 with gusts reaching 18 mph. Orlando surges to 94°F Sunday and 95°F by Tuesday July 07, with Monday’s gusts hitting 20 mph. Miami’s rain chance rises to 55% Sunday at 88°F, holding at 55% Monday before easing to 38% Tuesday at 89°F.