Question

An object that is falling or vertically projected into the air has it height, in feet, above the ground given by the formula below, where s is the height in feet, Vo is the original velocity of the object, in feet per second, t is the time the object is in motion in seconds, and So is the height, in feet, from which the object is dropped or projected. The figure shows that a ball is thrown straight up from a rooftop at an original velocity of 24 feet per second from a height of 96 feet. The ball misses the rooftop on its way down. How high above the ground will the ball be 2 seconds after thrown?
s=16t^2+VottSo. Please help, I'm stuck
fix the typo, and plug in the numbers

s = 96 + 24t - 16t^2
when t=2,
s(2) = 96+48-64 = 80

Related Questions

The height h after t seconds of a falling object starting at t=0 at height x feet is h(t)=x−16t^2 (f... From a platform, an object is projected vertically into the air. Its height, metres above ground l... From a platform an object is projected vertically into air. Its height, h metres above ground level... The height of an object that is projected can be modeled using the formula h=−16t2+v0t+h0, where h i...