Question

A hawk drops its prey from a certain hieght above the ground. The height, h metres, of the prey can be modelled by h=4+11t-3t², where t is the time in seconds after it is dropped by the hawk.
1. At what height above the ground does the hawk drops its prey?
2. at what time will the prey fall onto the ground?

Answer 1

Answer:

1. Height $= 7,848$ m

2. Time taken = 4 s

Step-by-step explanation:

$h = 4 + 11t - 3t^2$

When the prey falls onto the ground, h=0

∴ $3t^2 - 11t - 4 = 0$

$=> 3t^2 - 12t + t - 4 = 0$

$=>(t - 4)(3t + 1) = 0$

3t + 1 ≠ 0 [since this makes t negative]

∴ t = 4 s

$h = \frac{1}{2} g t^2$ [It was a free fall - zero initial velocity]

∴ $h = \frac{1}{2} . 981 . 4^2$ $= 7,848$ m