The projectile is fired at an angle of 45 degree to the horizontal with an initial speed of 100 feet per second. Show work.
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To solve this problem, we need to break down the initial velocity into its horizontal and vertical components. Let's assume that the positive x-direction is parallel to the ground, and the positive y-direction is vertically upward.
Given:
Initial speed (vi) = 100 feet per second
Launch angle (θ) = 45 degrees
Step 1: Finding the horizontal component of velocity (Vx):
The horizontal component of velocity (Vx) remains constant throughout the projectile's motion.
Vx = vi * cos(θ)
Plugging in the values:
Vx = 100 ft/s * cos(45°)
Vx = 100 ft/s * (√2/2)
Vx = 50√2 ft/s
Step 2: Finding the vertical component of velocity (Vy):
The vertical component of velocity (Vy) changes due to gravity. At the highest point, Vy equals zero.
Vy = vi * sin(θ)
Plugging in the values:
Vy = 100 ft/s * sin(45°)
Vy = 100 ft/s * (√2/2)
Vy = 50√2 ft/s
Step 3: Analyzing the motion in the x-direction:
As there are no horizontal forces acting on the projectile (assuming no air resistance), its horizontal velocity remains constant.
Step 4: Analyzing the motion in the y-direction:
The projectile is subject to the downward acceleration due to gravity (g ≈ 32 ft/s²). The time taken to reach the highest point can be determined using the vertical component of velocity.
Vy = uy + gt
0 = 50√2 ft/s + (-32 ft/s²) * t
Solving for t, we find:
t = (50√2 ft/s) / (32 ft/s²)
t ≈ 1.25 seconds
Step 5: Finding the maximum height (H):
We can use the time taken to reach the highest point to calculate the maximum height attained by the projectile.
H = uy * t + (1/2) * (-32 ft/s²) * t²
H = 50√2 ft/s * 1.25 s + (1/2) * (-32 ft/s²) * (1.25 s)²
Simplifying the equation:
H = 62.5√2 ft + (-20 ft/s²) * 1.5625 s²
H ≈ 78.125 ft
So, by following these steps, we can find that the maximum height reached by the projectile is approximately 78.125 feet.