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

Answer:

(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.

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:

$x = x_{o}+v_{o,x} \cdot t$

$y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

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 $x_{o} = 0\,m$, $y_{o} = 80\,m$, $v_{o,x} = 360\,\frac{m}{s}$, $v_{o,y} = 0\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$, 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:

$80 - 4.094\cdot t^{2} = 0$

$t = \sqrt{\frac{80}{4.094} }\,s$

$t \approx 4.420\,s$

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.