h(t)= -204.8t sq 256t - find maximum height of jump where h(t) is height in inches and t is seconds after beginning jump
h(t)= -204.8t sq 256t -
makes no sense to me
Please clarify your equation
To find the maximum height of the jump, we need to determine the vertex of the parabolic function h(t) = -204.8t^2 + 256t.
The vertex of a parabola in the form of h(t) = at^2 + bt + c is given by the formula t = -b / (2a).
In our case, a = -204.8 and b = 256. Plugging these values into the formula, we get:
t = -256 / (2 * -204.8)
= -256 / -409.6
= 0.625
Therefore, the time at which the ball reaches its maximum height is 0.625 seconds.
Now, to find the maximum height, substitute this time back into the original equation:
h(0.625) = -204.8 * (0.625)^2 + 256 * 0.625
= -204.8 * 0.390625 + 160
= -79.6875 + 160
= 80.3125
Hence, the maximum height of the jump is approximately 80.31 inches.