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dmitriy555 [2]
2 years ago
11

Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas

e show work!
Physics
1 answer:
Westkost [7]2 years ago
8 0

Answer:

<em>The range is 15.15 m and the time in the air is 1.01 s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

To calculate the time the object takes to hit the ground, we use the equation below:

\displaystyle t=\sqrt{\frac{2h}{g}}

The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:

\displaystyle d=15\cdot\sqrt{\frac  {2*5}{9.8}}=15*1.01

d = 15.15 m

The time in the air is:

\displaystyle t=\sqrt{\frac{2*5}{9.8}}

t = 1.01 s

The range is 15.15 m and the time in the air is 1.01 s

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