a projectile is fired from the ground with an initial horizontal velocity of 8 meters per second. its initial vertical velocityi
s 6 meters per second. whats is the "total/resultant" velocity of the projectile at its peak?
1 answer:
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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:
- 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:
- 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:
- 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:
- Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:
- 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.