Question

Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees. Using your knowledge of kinematics equations, determine whether this would be sufficient to deliver the payload to the drop site.

Answer 1

Answer:

Please see below as the answer is self-explanatory.

Explanation:

• We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
• Both movements are independent each other, due to they are perpendicular.
• In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
• Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

       $v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)$

• Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

       $t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)$

• In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
• Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

       $\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)$

• In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

       $v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)$

• Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

       $\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)$

• Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.