The City of Brownsville Parks and Recreation Department is inviting residents and visitors to celebrate Independence Day with an evening of family-friendly entertainment, community spirit, and a spectacular fireworks show at Brownsville Sports Park.
The annual Fourth of July celebration will take place on Friday, July 4, and promises a full lineup of activities for all ages. Guests can enjoy live music, local vendors, food, and a classic car show before the night concludes with one of Brownsville's favorite holiday traditions—a dazzling fireworks display.
The celebration begins at 5 p.m. with the opening of the tailgate area, giving families and friends an opportunity to gather and enjoy the festive atmosphere. Starting at 6 p.m., attendees can explore the Stars & Stripes Vendor Market, browse unique merchandise, admire classic vehicles, and enjoy live entertainment throughout the evening.
The highlight of the night will be the Fireworks Spectacular at 9:20 p.m., lighting up the sky in celebration of America's birthday.
The event provides an opportunity for the community to come together, celebrate the nation's independence, and create lasting memories with family and friends.
Event Details
Friday, July 4
Brownsville Sports Park
1000 Sports Park Blvd.
5:00 p.m. – Tailgate Area Opens
6:00 p.m. – Live Music, Stars & Stripes Vendor Market, Classic Car Show, and Family Activities
9:20 p.m. – Fireworks Spectacular
Residents are encouraged to bring lawn chairs, gather their family and friends, and join the City of Brownsville for an evening of patriotic fun, entertainment, and community celebration.
Source: City of Brownsville
