A ball is thrown into the air with an upward velocity of 48 ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+48t+6. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

a. 3 s; 6 ft
b.1.5 s; 54 ft
c. 1.5 s; 114 ft
d. 1.5 s; 42 ft

Question

A ball is thrown into the air with an upward velocity of 48 ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+48t+6.  In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

a. 3 s; 6 ft
	b.1.5 s; 54 ft
	c. 1.5 s; 114 ft
	d. 1.5 s; 42 ft

anonymous · Answer

recall that the velocity of the ball at maximum height is 0.

for velocity function. Take the derivative of the height function.
$$\frac{ d(h) }{ dt } = -32t + 48$$

solve the velocity function for 0 and find t than substitute that back into the height function the maximum height.

whpalmer4 · Answer

Or, notice that this is a parabola, written in form $y = ax^2+bx+c$, so the x coordinate of the vertex (which will be the maximum) is simply -b/2a.  Then evaluate the function at that x value to get the maximum height.

whpalmer4 · Answer

x -> t
y -> h
in this case