Question

Air drop A plane traveling horizontally at 80 m/s over flat ground at an elevation of 3000 m releases an emergency packet. The trajectory of the packet is given by
x = 80t, y = −4.9t² + 3000, t ≥ 0
where the origin is the point on the ground directly beneath the plane at the moment of the release (see figure). Graph the trajectory of the packet and find the coordinates of the point where the packet lands.

Accepted Answer

Step 1: Understand the given parametric equations for the trajectory of the packet: $$x = 80t$$ and $$y = -4.9t$$^{2}$$ + 3000$$, where $$t \geq 0. $$Here, $$x$$ represents the horizontal distance from the origin, and $$y$$ represents the height above the ground. Step 2: To graph the trajectory, create a table of values by choosing several values of $$t ($$starting from 0) and calculating the corresponding $$x$$ and $$y$$ coordinates using the given equations. Plot these points on the coordinate plane. Step 3: To find the point where the packet lands, determine when the packet reaches the ground, i.e., when $$y = 0. $$Set the equation $$-4.9t^{2} + 3000 = 0$ and solve for t$. Step 4: Once you find the time t$ when the packet hits the ground, substitute this value back into the horizontal position equation x = 80t$ to find the horizontal distance from the origin where the packet lands. Step 5: The coordinates of the landing point are then (x$$, 0)$$, where x$ is the horizontal distance calculated in Step 4. This point represents where the packet touches the ground.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Parametric Equations

Projectile Motion

Finding the Landing Point