Question

A ball is thrown upward from a height of 15 feet with an initial upward velocity of 5 feet per second. Use the formula b(t)= -16t^2 + 5t + 15 to find the time it takes for the ball to reach the maximum height of the ball. Round to the nearest tenth
A. 0.2 seconds

B. 0.1 seconds

C. 0.3 seconds

D. 0.9 seconds​​

Answer 1

Answer:

A) 0.2 seconds

Step-by-step explanation:

$b(t)=-16t^2+5t+15$

$v(t)=-32t+5$ <-- Take the derivative

$0=-32t+5$ <-- Set equal to 0

$-5=-32t$

$\frac{5}{32}=t$

$t=\frac{5}{32}$

$t\approx0.2$

Therefore, it will take the ball about 0.2 seconds to reach its maximum height (which happens to be about 15.4 feet BTW).

Alternatively, another non-calculus approach is the fact that the parabola opens downward, and as such, the maximum of $b(t)$ occurs at the vertex where $t=-\frac{b}{2a}$, therefore $t=-\frac{5}{2(-16)}=\frac{-5}{-32}=\frac{5}{32}\approx0.2$

Answer 2

Answer:

0.2 seconds

Step-by-step explanation:

The graph of b(t) = -16t^2 + 5t + 15 is a parabola opening downward.  The maximum value of b(t) occurs at the vertex.  The t-coordinate of the vertex is t = -5 / (2(-16)) = 5/32 = 0.15625 sec ≈ 0.2 sec.