An object is dropped from a tower, 1600 ft above the ground. the object's height above the ground t seconds after the fall is s(t)=1600-16(t^2). determine the velocity and acceleration of the object the moment it reaches the ground.

You'll need the time that the object will reach the ground, which can be obtained by solving for s(t) = 0:1600 - 16t^2 = 01600 = 16t^2100 = t^2t = 10Once you have this time, you can plug it into the velocity and acceleration function, which the first and second derivative of s(t), respectively.v(t) = s'(t) = -32ta(t) = -32 (as expected, free fall acceleration due to gravity is constant)Then v(10) = -320 ft/sAnd a(10) = -32 ft/s^2
First determine how long it takes for the object to hit the ground.Set s(t) =1600 - 16t^2 = 0So t = 10s would be the time taken.Take the first derivative of s(t) to find the velocity-time relationship:s'(t) = -32tSo s(10) = -320ft/s... This is your velocity.I think the acceleration at all points close to the earth is roughly constant, -9.81m/s^2 = -32ft/s^2
So velocity is first derivative.