Answer:
[tex]d = 13.3 \times 0.61 = 8.06 m[/tex]
Explanation:
If a marble is shot straight up in the air then the maximum height above the ground is given as 9 m
so here we will have
[tex]mgh = \frac{1}{2}mv^2[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2(9.8)(9)} = 13.3 m/s[/tex]
now the same marble is shot horizontally so the speed will be same
now the height from which it is shot is given as 1.8 m
so time taken by it to reach the ground is given as
[tex]y = \frac{1}{2}gt^2[/tex]
[tex]1.8 = \frac{1}{2}(9.8) t^2[/tex]
[tex]t = 0.61 s[/tex]
now the distance moved by the marble is given as
distance = speed x time
[tex]d = 13.3 \times 0.61 = 8.06 m[/tex]
The range of the marble if it is fired horizontally is 8.10 meters.
Given the following data:
To calculate the range of the marble if it is fired horizontally:
We would determine the vertical speed of the marble fired vertically by the spring-loaded gun through this formula:
[tex]V=\sqrt{2gh} \\\\V=\sqrt{2 \times 9.8 \times 9.0}\\\\V=\sqrt{176.4}[/tex]
V = 13.28 m/s.
Next, we would calculate the time taken to reach the maximum height:
[tex]t=\sqrt{\frac{2h}{g} } \\\\t=\sqrt{\frac{2 \times 1.8}{9.8} }\\\\t=\sqrt{0.3674}[/tex]
Time, t = 0.61 seconds.
For the marble's range:
[tex]Range = 13.28 \times 0.61[/tex]
Range = 8.10 meters.
Read more on range here: https://brainly.com/question/5870402
