a projectile is fired from the vertical tube mounted on the vehicle which is travelling at the constant speed of 30kph. the projectile leaves the tube with a velocity of 20m/s relative to the tube.neglecting air resistance.
what is the question?
a. det. distance traveled by vehicle during the flight of projctile where the projectile will land on the vehicle tube location
b. det. the time of flight until it will land back on the tube location
c. max height of projectile
To analyze the motion of the projectile, we need to break the problem into two components: the horizontal and vertical motions.
First, let's consider the horizontal motion. Since the vehicle is traveling at a constant speed of 30 kph, we can assume that there are no horizontal forces acting on the projectile. This means that the horizontal velocity of the projectile will also be constant at 30 kph, or 8.33 m/s.
Next, let's focus on the vertical motion. We know that the projectile leaves the tube with a velocity of 20 m/s relative to the tube. Assuming the tube is vertical, the initial vertical velocity of the projectile will also be 20 m/s.
Now, let's determine the time it takes for the projectile to reach its highest point. To do this, we can use the fact that the vertical motion of the projectile is governed by the laws of free fall. The initial vertical velocity (20 m/s) will gradually decrease due to the acceleration of gravity (-9.8 m/s^2) until it reaches zero at the highest point of the trajectory. At this point, the projectile will start to fall downward.
We can use the following equation to find the time it takes for the projectile to reach its highest point:
v = u + at
In this equation, v represents the final vertical velocity (0 m/s), u represents the initial vertical velocity (20 m/s), a represents the vertical acceleration (-9.8 m/s^2), and t represents the time.
By substituting the known values into the equation, we get:
0 = 20 - 9.8t
Solving for t, we find:
t = 2.04 seconds
So, it will take approximately 2.04 seconds for the projectile to reach its highest point.
Now, let's determine the vertical distance traveled by the projectile. We can use the equation for displacement in vertical motion:
s = ut + (1/2)at^2
In this equation, s represents the vertical displacement, u represents the initial vertical velocity (20 m/s), a represents the vertical acceleration (-9.8 m/s^2), and t represents the time (2.04 seconds).
By substituting the known values into the equation, we get:
s = (20)(2.04) + (1/2)(-9.8)(2.04)^2
Simplifying the equation, we find:
s = 20.4 - 9.82
Therefore, the vertical distance traveled by the projectile is approximately 10.2 meters.
Finally, to determine the total distance traveled by the projectile, we can use the fact that the horizontal and vertical motions are independent of each other. Since the horizontal velocity is constant at 8.33 m/s, and the time of flight from the highest point to the ground is approximately 4.08 seconds (twice the time to reach the highest point), we can calculate the horizontal distance traveled using the equation:
d = v * t
In this equation, d represents the horizontal distance traveled (30 kph or 8.33 m/s multiplied by 4.08 seconds).
By substituting the known values into the equation, we get:
d = 8.33 * 4.08
Therefore, the horizontal distance traveled by the projectile is approximately 34 meters.
In conclusion, neglecting air resistance, a projectile fired from a vertical tube mounted on a vehicle traveling at a constant speed of 30 kph will have a constant horizontal velocity of 8.33 m/s. The projectile will reach its highest point in approximately 2.04 seconds, traveling a vertical distance of approximately 10.2 meters. The projectile will then travel a horizontal distance of approximately 34 meters before reaching the ground.