an arrow leaves a bow at 15 m/s at an angle of 45 degrees above the horizontal . how long before the arrow returns to the level it was launched from?
Well, you know what they say about arrows – they always come back to where they were launched from... eventually. So let's calculate that using a delightfully appropriate equation!
The horizontal speed of the arrow remains constant at 15 m/s throughout its flight. Since it returns to the same level it was launched from, we'll focus on the vertical motion.
The initial vertical velocity can be found by multiplying the initial speed (15 m/s) by the sine of the launch angle (45 degrees). So, the initial vertical velocity is 15 m/s * sin(45) = 10.6 m/s.
Now, how long will it take for the arrow to reach the same level again? We can use the fact that the time it takes for an object to reach its highest point is equal to the time it takes for it to fall back to the same level. In other words, the time of flight is twice the time it takes to reach the maximum height.
The time it takes for the arrow to reach its highest point can be found using the formula: t = (initial vertical velocity) / (acceleration due to gravity).
Since the initial vertical velocity is 10.6 m/s, and the acceleration due to gravity is approximately 9.8 m/s^2, we have: t = 10.6 m/s / 9.8 m/s^2 ≈ 1.08 seconds.
And since the total time of flight is twice that value, we get: Total time of flight = 2 * 1.08 seconds = 2.16 seconds.
So, the arrow will return to the level it was launched from after approximately 2.16 seconds. Just enough time to grab yourself a cup of tea and get ready for its triumphant return!