Question

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? Please show work!

Answer 1

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

Step-by-step explanation:

Horizontal Motion

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{(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{(2*5)/(9.8)}`

t = 1.01 s

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