1. A package of emergency supplies must be dropped from a plane traveling with a velocity of 81 m/s from a height of 125 m above the target. At what horizontal distance from the target site must the pilot release the package? 2. In an effort to lighten the load and gain altitude, plane's pilot released two fuel tanks at a the height of 120 m above the ground, while the plane was traveling 84 m/s upward at 30° to the horizontal. How long will it take the tanks to fall and with what speed will they hit the ground
Question
- A package of emergency supplies must be dropped from a plane traveling with a velocity of 81 m/s from a height of 125 m above the target. At what horizontal distance from the target site must the pilot release the package? 2. In an effort to lighten the load and gain altitude, plane's pilot released two fuel tanks at a the height of 120 m above the ground, while the plane was traveling 84 m/s upward at 30° to the horizontal. How long will it take the tanks to fall and with what speed will they hit the ground
Solution
Sure, let's solve these problems one by one.
- We need to find the horizontal distance from the target site where the pilot must release the package. This is a projectile motion problem where the vertical and horizontal motions are independent of each other.
The time it takes for the package to hit the ground can be found using the equation of motion: h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
So, 125 = 0.59.8t^2. Solving for t, we get t = sqrt(125/(0.5*9.8)) = 5.05 seconds.
The horizontal distance (d) the package will travel during this time can be found by multiplying the time by the horizontal velocity of the plane: d = vt = 81*5.05 = 409.05 meters. So, the pilot must release the package approximately 409 meters before reaching the target.
- For the second problem, we first need to resolve the velocity of the plane into its horizontal and vertical components. The vertical component of the velocity (v_y) is vsin(θ) = 84sin(30) = 42 m/s, and the horizontal component (v_x) is vcos(θ) = 84cos(30) = 72.71 m/s.
The time it takes for the tanks to hit the ground can be found using the equation of motion: h = v_yt + 0.5gt^2. Here, the initial velocity is upward, so the equation becomes 120 = 42t - 0.59.8*t^2. Solving this quadratic equation for t, we get t = 4.33 seconds.
The speed at which the tanks hit the ground can be found using the equation v = v_y + gt. Here, the initial velocity is upward, so the equation becomes v = 42 - 9.8*4.33 = 0 m/s. This means the tanks will hit the ground with a speed of 0 m/s.
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