Question

An object is dropped from a tower, 1024 ft above the ground. The object's height above ground i seconds after the fall is s(t) = 1024 - 1612. Determine the velocity and acceleration of the object the moment it reaches the ground. ft/s. The velocity of the object the moment it reaches the ground is (Simplify your answer.) The acceleration of the object the moment it reaches the ground is fus2 (Simplify your answer.)

Answer 1

We are quien the following in the question : Oljicts height at t seconds is given by s(t) = 1024 -167 Time when object reaches ground -) s(t) = 0 1024 -16+20 = 16t² = 1024 - 64 16 t² = 1024 = t = √64 18 Since, time cannot be regature =) t = 8 second The velocity is quien by vct) d [s(t) dt d [1024-16+ ² dt - 32t Putting it = 8 sec

vct) - -3218) - 256 ft/sec The acceleration is quien hay alt) od [u(t)] at d [-32] dt -32 / tec? The velocity of elegiet is - 256 $t/ sxc ard acceleration is - 32 ft/ seca. The regature sign shows the opposite diruction