An object has been launched vertically upward from a building. The height of the object relate to time can be found by h(t)= -16t^2 +48t+960. (pls use this when solving the problem thanks)
A. What was the speed when the object was launched?
B. How tall was the building?
C. What is the maximum height the object could reach?
D. When will the object reach its maximum height?
E. When will the object hit the ground?
Thanks!
+129643 A = 48 ft/ s
B = 960 ft
C Find the t coordinate of the vertex = - 48 / (2 * -16) = 48/32 = 3/2 sec
Plug this back into the function
-16(3/2)^2 + 48(3/2) + 960 = 996 ft = maax height
D ( from C ) = 3/2 sec
E The object will hit to ground when h = 0....so
-16t^2 + 48t + 960 = 0 divide through by -16
t^2 - 3t - 160 = 0 complete the square on t
t^2 -3t + 9/4 = 160 + 9/4
(t - 3/2)^2 = 649/4 take both roots and we have that
t = sqrt (694)/2 + 3/2 = 14.67 sec or t = -sqrt (694/)/2 +3/2 = -11.67 sec
Take the positive answer = 14.67 sec
