A baseball is thrown upward from ground level with an initial velocity of 48 feet per second, and its height h in feet is given by , where t is the time in seconds.

Question

A baseball is thrown upward from ground level with an initial velocity of 48 feet per second, and its height h in feet is given by  ,   where t is the time in seconds.
At what time does the ball reach 10 feet?  Round to the nearest tenth.

Step by step please

Steve · Answer

h(t) = 48t - 16t^2 so, plug in 10 for h and solve for t: 10 = 48t - 16t^2 8t^2 - 24t + 5 = 0 t = (6±√26)/4 or t=0.225s (on the way up) t=2.774s (on the way back down)