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
and ... ?
To solve this problem, we can break it down into two components: the horizontal motion of the vehicle and the vertical motion of the projectile.
First, let's analyze the horizontal motion. We are given that the vehicle is traveling at a constant speed of 30 km/h (or 8.33 m/s). Since there is no external force acting on the projectile horizontally, its horizontal velocity remains constant throughout its motion. Therefore, the horizontal velocity of the projectile is 8.33 m/s.
Now, let's focus on the vertical motion. The projectile leaves the tube with a velocity of 20 m/s relative to the tube. Since there are no external forces acting on the projectile vertically (neglecting air resistance), we can use the equations of motion under constant acceleration.
The initial vertical velocity of the projectile, u, is 20 m/s. The final vertical velocity, v, is unknown. The acceleration, a, is equal to the acceleration due to gravity, -9.8 m/s^2 (negative because we consider upward as positive).
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time, we can solve for the time it takes for the projectile to reach its peak height. Since the projectile is fired vertically upward, it will reach its peak height when its vertical velocity becomes zero. Plugging in the values, we get:
0 = 20 - 9.8t
9.8t = 20
t = 20 / 9.8
t ≈ 2.04 seconds
Therefore, it takes approximately 2.04 seconds for the projectile to reach its peak height.
Next, let's find the maximum height reached by the projectile. We can use the equation for vertical displacement, s, under constant acceleration:
s = ut + (1/2)at^2
Plugging in the values, we get:
s = (20)(2.04) + (1/2)(-9.8)(2.04)^2
s ≈ 20.4 - 20.1
s ≈ 0.3 meters
Therefore, the maximum height reached by the projectile is approximately 0.3 meters.
Finally, we can determine the total horizontal distance traveled by the projectile. Since there is no horizontal acceleration, we can use the equation:
distance = speed * time
Plugging in the values, we get:
distance = 8.33 * 2.04
distance ≈ 16.97 meters
Therefore, the total horizontal distance traveled by the projectile is approximately 16.97 meters.