. A hawk drops its prey from a certain height above the ground.The height, h metres, of the prey can be modelled by h = 4 + 11 t

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. A hawk drops its prey from a certain height above the ground.The height, h metres, of the prey can be modelled by h = 4 + 11 t

– 3t2, where t is the time in seconds after it is dropped by the hawk. At what height above the ground does the hawk drop its prey? At what time will the pray fall onto the ground?

Mathematics

1 answer:

Leona [12.6K] · Answer 1 · 2026-04-08T16:17:34+03:00

Answer:

The hawk releases the prey from a height of 4 meters.

It takes the prey 4 seconds to reach the ground.

Step-by-step explanation:

The equation gives insights about the height of the prey at any time starting from the moment it is dropped. Thus, to determine the drop height, we evaluate the expression at time equals zero (the drop moment). This answers the first question:

$h=4+11t-3t^2\\h=4+11\,(0)-3\,(0)^2\\h=4\, \,meters$

To ascertain when the prey touches the ground, we set "h" to zero (height of zero) and solve for "t".

This results in a quadratic equation that can be solved via the quadratic formula:

$h=4+11t-3t^2\\0=4+11t-3t^2\\-3t^2+11t+4=0\\t=\frac{-11+-\sqrt{11^2-4\,(-3)(4)} }{2\,(-3)} \\t=\frac{-11+-\sqrt{121+48} }{-6} \\t=\frac{-11+-13 }{-6} \\t= 4\,\,and \,\, t=-\frac{1}{3}$

Since negative time values are impractical, we select the positive 4 (4 seconds)