Question

A projectile is fired from a height of 80 M above sea level, horizontally with a speed of 360 M / S, calculate: The time it takes for the projectile to reach the water. The Horizontal scope. The height that remains to descend after 2 seconds of being launched.

Answer 1

(a) The projectile takes approximately 4.420 seconds to reach the water, (b) The horizontal scope of the projectile is 1591.2 meters, (c) The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Step-by-step explanation:

The projectile experiments a parabolic motion, where horizontal speed remains constant and accelerates vertically due to the gravity effect. Let consider that drag can be neglected, so that kinematic equation are described below:

`x = x_(o)+v_(o,x) \cdot t`

`y = y_(o) + v_(o,y)\cdot t +(1)/(2)\cdot g \cdot t^(2)`

Where:

`x_(o)`,
`y_(o)` - Initial horizontal and vertical position of the projectile, measured in meters.

`v_(o,x)`,
`v_(o,y)` - Initial horizontal and vertical speed of the projectile, measured in meters per second.

`t` - Time, measured in seconds.

`g` - Gravitational acceleration, measured in meters per square second.

`x`,
`y` - Current horizontal and vertical position of the projectile, measured in meters.

Given that
`x_(o) = 0\,m`,
`y_(o) = 80\,m`,
`v_(o,x) = 360\,(m)/(s)`,
`v_(o,y) = 0\,(m)/(s)` and
`g = -9.807\,(m)/(s^(2))`, the kinematic equations are, respectively:

`x = 360\cdot t`

`y = 80-4.094\cdot t^(2)`

(a) If
`y = 0\,m`, the time taken for the projectile to reach the water is:

`80 - 4.094\cdot t^(2) = 0`

`t = \sqrt{(80)/(4.094) }\,s`

`t \approx 4.420\,s`

The projectile takes approximately 4.420 seconds to reach the water.

(b) The horizontal scope is the horizontal distance done by the projectile before reaching the water. If
`t \approx 4.420\,s`, the horizontal scope of the projectile is:

`x = 360\cdot (4.420)`

`x = 1591.2\,m`

The horizontal scope of the projectile is 1591.2 meters.

(c) If
`t = 2\,s`, the height that remains to descend is:

`y = 80-4.094\cdot (2)^(2)`

`y = 63.624\,m`

The remaining height to descend after 2 seconds of being launched is 63.624 meters.