An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = -4.9t^2 + 19.6t + 58.8, where s is in meters. How high will the object be after 2 seconds?

194. 04 feet
117.6 feet
78.4 feet
1.96 feet

pls help 20 points tyy

Planet earth has a mass of 5.9726x10²⁴ kg.then: 5.9726x10²⁴ kg=100% of the total earth mass.if the water´s mass=5x10-4 % of the total mass, we can solve this problem by the rule of three.5.9726x10²⁴ %*10⁻⁴ %x=(5.9726x10²³ kg)(5*10⁻⁴  %) / 100 %=2.9863x10¹⁹ kganswer:   the mass of the planet´s water would be   2.9863 x 10¹⁹ kg.

D- dialation and reflection