A ball is thrown vertically upward with an initial velocity of 64 feet per second. The distance(in feet) of the ball from the ground after t seconds is s=64t-16t.
At what time will the ball strike the ground?
For what time t is the ball more than 28 feet above the ground.
The ball will strike the ground when t is _____ seconds.
The ball is more than 28 feet above the ground for the time t when ____<t<____.
2 answers
you should proably first simplify 64t-16t = 48t. Then, 48t=s. Plug it in, see what you get.
s = 64t - 16t^2
s=0 at t= 0,2
ball is more than 28 ft high when
64t - 16t^2 > 28
-16t^2 + 64t - 28 > 0
-4(2t-1)(2t-7) > 0
1/2 < t < 7/2
s=0 at t= 0,2
ball is more than 28 ft high when
64t - 16t^2 > 28
-16t^2 + 64t - 28 > 0
-4(2t-1)(2t-7) > 0
1/2 < t < 7/2
1 answer