an artillery piece is pionted upward at an angle of 35degree with respect to the horizontal and fire a projectile with a muzzle velocity 200meter per second .if air resistance is negligible to what height will the projectile rise.
h=vo2sin2theta\2g
h=(200)2*sin 2 35\19.6
h=40000*0.36\19.6
h=672.4
Vo = 200m/s[35o].
Yo = 200*sin35 = 114.7 m/s.
Y^2 = Yo^2 + 2g*h.
Y = 0.
g = -9.8 m/s^2.
h = ?
To find the height the projectile will rise, we can use the equations of projectile motion.
Step 1: Break down the initial velocity into its horizontal and vertical components.
The horizontal component (Vx) remains constant throughout the motion and is given by:
Vx = V * cos(θ)
where V is the muzzle velocity and θ is the launch angle.
The vertical component (Vy) changes over time due to the effect of gravity and is given by:
Vy = V * sin(θ)
Step 2: Determine the time it takes for the projectile to reach the peak of its trajectory.
The time it takes for an object to reach its peak height can be found using the equation:
t = Vy / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Step 3: Calculate the maximum height (h) reached by the projectile.
The maximum height h can be determined using the equation:
h = (Vy^2) / (2 * g)
Step-by-step solution:
Given:
Launch angle (θ) = 35°
Muzzle velocity (V) = 200 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Step 1:
Vx = V * cos(θ)
Vx = 200 * cos(35°)
Vx ≈ 200 * 0.8192
Vx ≈ 163.84 m/s
Vy = V * sin(θ)
Vy = 200 * sin(35°)
Vy ≈ 200 * 0.5736
Vy ≈ 114.72 m/s
Step 2:
t = Vy / g
t = 114.72 / 9.8
t ≈ 11.712 s (rounded to three decimal places)
Step 3:
h = (Vy^2) / (2 * g)
h = (114.72^2) / (2 * 9.8)
h ≈ 660.636 m (rounded to three decimal places)
Therefore, the projectile will rise to a height of approximately 660.636 meters.
To find the height the projectile will reach, we can use the principles of projectile motion. We need to consider the initial velocity, launch angle, and the effect of gravity.
The first step is to break down the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.
Given:
Initial velocity (v) = 200 m/s
Launch angle (θ) = 35 degrees
To find the horizontal component (vx) of the velocity, we use the equation:
vx = v * cos(θ)
Substituting the given values:
vx = 200 m/s * cos(35°)
To find the vertical component (vy) of the velocity, we use the equation:
vy = v * sin(θ)
Substituting the given values:
vy = 200 m/s * sin(35°)
Now we need to determine the time taken for the projectile to reach its maximum height. At the maximum height, the vertical velocity becomes zero since the projectile momentarily stops before falling back down. We can use the equation:
vy = vy0 + at
At the highest point, vy = 0, so the equation becomes:
0 = vy0 + (-g)t
Where:
vy0 is the initial vertical component of velocity, which is vy.
g is the acceleration due to gravity (approximately 9.8 m/s²).
t is the time taken to reach the maximum height.
Solving for t:
0 = (200 m/s * sin(35°)) - (9.8 m/s² * t)
Using basic algebra, we can rearrange the equation to solve for t:
t = (200 m/s * sin(35°)) / 9.8 m/s²
Now we have determined the time taken to reach the maximum height.
To find the height (h), we can use the equation:
h = vy0 * t + (1/2) * (-g) * t²
Substituting the values:
h = (200 m/s * sin(35°)) * [(200 m/s * sin(35°)) / 9.8 m/s²] + (1/2) * (-9.8 m/s²) * [(200 m/s * sin(35°)) / 9.8 m/s²]^2
Simplifying the equation, we can calculate the height the projectile will reach.