Question

12.
The height (in meters) at any time t (in seconds)
of a ball thrown vertically varies according to
equation h(t) = -16t2 + 256t. How long after
in seconds the ball reaches the highest point.
​

Answer 1

Answer:

8 s

Explanation:

This is a maximisation problem,

In such cases to get the maximum value of h

we have to find the value of t by the equation, dh/dt = 0

ie $\frac{d}{dt} (-16t^{2} + 256t) = 0\\-32t+256 = 0\\t = 256/32 = 8 s$

the ball will reach the highest point after 8 s

To find the maximum height , insert t = 8s

max height hmax =  - 16 × 8 ×8 + 256 × 8 = 1024m

Answer 2

Answer:16

Explanation:

shift solve