Question

a helicopter is rising vertically above level ground at a constant speed of 8.7 m/s. at a height of 10.5 m above the ground, a parcel is thrown out of the helicopter. relative to the helicopter, the parcel is thrown with a velocity of 7.5 m/s at an angle of 27° above the horizontal. how far from the point on the ground directly beneath the helicopter does the parcel land?

Explanation:

Step1: Analyze vertical - motion of the parcel

The initial vertical position of the parcel $y_0 = 10.5$ m, the initial vertical velocity of the parcel $v_{0y}=8.7 + 7.5\sin(27^{\circ})$ m/s, and the acceleration $a=-g=- 9.8$ m/s². When the parcel lands on the ground $y = 0$. We use the equation $y - y_0=v_{0y}t+\frac{1}{2}at^{2}$, so $0 - 10.5=(8.7 + 7.5\sin(27^{\circ}))t-\frac{1}{2}\times9.8t^{2}$.
First, calculate $7.5\sin(27^{\circ})\approx7.5\times0.454 = 3.405$. Then $v_{0y}=8.7 + 3.405=12.105$ m/s. The equation becomes $- 10.5 = 12.105t-4.9t^{2}$, or $4.9t^{2}-12.105t - 10.5 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here $a = 4.9$, $b=-12.105$, $c=-10.5$.
$t=\frac{12.105\pm\sqrt{(-12.105)^{2}-4\times4.9\times(-10.5)}}{2\times4.9}=\frac{12.105\pm\sqrt{146.53+205.8}}{9.8}=\frac{12.105\pm\sqrt{352.33}}{9.8}=\frac{12.105\pm18.77}{9.8}$.
We take the positive root $t=\frac{12.105 + 18.77}{9.8}=\frac{30.875}{9.8}\approx3.15$ s.

Step2: Analyze horizontal - motion of the parcel

The initial horizontal velocity of the parcel $v_{0x}=7.5\cos(27^{\circ})$. Calculate $7.5\cos(27^{\circ})\approx7.5\times0.891 = 6.6825$ m/s.
In horizontal - motion (where there is no acceleration, $a_x = 0$), we use the equation $x=v_{0x}t$. Substitute $v_{0x}=6.6825$ m/s and $t = 3.15$ s. Then $x=6.6825\times3.15\approx21.15$ m.

Answer:

Approximately $21.2$ m