Choice D is correct. An equation representing the height above ground $h$ , in meters, of a softball $t$ seconds after it is launched by a machine from ground level can be written in the form $h = - a {(t - b)}^{2} + c$ , where $a$ , $b$ , and $c$ are positive constants. In this equation, $b$ represents the time, in seconds, at which the softball reaches its maximum height of $c$ meters above the ground. It's given that this softball reaches a maximum height of $51.84$ meters above the ground at $1.8$ seconds; therefore, $b equals 1.8$ and $c equals 51.84$ . Substituting $1.8$ for $b$ and $51.84$ for $c$ in the equation $h = - a {(t - b)}^{2} + c$ yields $h = - a {(t - 1.8)}^{2} + 51.84$ . It's also given that this softball hits the ground at $3.6$ seconds; therefore, $h equals 0$ when $t equals 3.6$ . Substituting $0$ for $h$ and $3.6$ for $t$ in the equation $h = - a {(t - 1.8)}^{2} + 51.84$ yields $0 = - a {(3.6 - 1.8)}^{2} + 51.84$ , which is equivalent to $0 = - a {(1.8)}^{2} + 51.84$ , or $0 = - 3.24 a + 51.84$ . Adding $3.24 a$ to both sides of this equation yields $3.24 a equals 51.84$ . Dividing both sides of this equation by $3.24$ yields $a equals 16$ . Substituting $16$ for $a$ in the equation $h = - a {(t - 1.8)}^{2} + 51.84$ yields $h = - 16 {(t - 1.8)}^{2} + 51.84$ . Therefore, $h = - 16 {(t - 1.8)}^{2} + 51.84$ represents the height above ground $h$ , in meters, of this softball $t$ seconds after it is launched.

Choice A is incorrect. This equation represents a situation where the maximum height is $3.6$ meters above the ground at $0$ seconds, not $51.84$ meters above the ground at $1.8$ seconds.

Choice B is incorrect. This equation represents a situation where the maximum height is $51.84$ meters above the ground at $0$ seconds, not $1.8$ seconds.

Choice C is incorrect and may result from conceptual or calculation errors.