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Answer :
Final answer:
In a two-dimensional projectile motion, an object's maximum height and horizontal distance traveled can enable us to determine the launch velocity. The vertical and horizontal velocities are calculated separately and then combined to find the launch velocity.
Explanation:
Firstly, we need to understand that the physics concept at play here is projectile motion, which means that an object is moving in the horizontal and vertical directions under the influence of gravity alone. In this scenario we are assuming that the object is landing at the point from which it was launched, and there is no air resistance. The maximum height observed is in the vertical direction, and the horizontal distance travelled is four times this height.
In projectile motion, the vertical and horizontal motions are independent of each other. Let's start with the vertical motion. The formula to calculate the maximum height, H in projectile motion when initial vertical velocity is u and gravity is g is: H = u^2/(2g). We know in this case that the maximum height H is 36.3 m, and g (acceleration due to gravity) is approximately 9.8 m/s^2. So we can rearrange the equation to find the initial vertical velocity: u = sqrt(2gH).
Moving on to the horizontal motion. The distance travelled in the horizontal direction is four times the maximum height, which gives a distance of 4 * 36.3 m = 145.2 m. Horizontal motion is at constant velocity, so the distance travelled is the horizontal velocity, v, times the time of flight, t; also, the total time of flight is the same for vertical motion which is 2u/g where u is the initial vertical velocity. Substituting these in we find the initial horizontal velocity: v = 145.2 m / (2u/g).
Finally, in projectile motion, the launch velocity is equal to the square root of the sum of the squares of the initial horizontal and vertical velocities. This gives the launch velocity as: sqrt(u^2 + v^2).
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