A projectile is launched at 5.4 m/s with an initial angle of 22 degrees from an initial height of 2.6 meters. Where will it land? How fast, and at what angle will it be going when it hits the ground?
Vhoriz=5.4*cos22
Vivertical=5.4 sin22
hf=hi+viv*time-4.9t^2
you hf=0,solve for time, it is quadratic.
vvfinal=5.4sin22-gt solve for vfinal
angle when landing:
arctan (vvfinal/vhoriz)
velocityfinal=sqrt(vverfinal^2+vhor^2)
To find out where the projectile will land, we need to analyze its motion and calculate the horizontal and vertical components of its velocity.
First, let's break down the given information:
Initial velocity (v₀) = 5.4 m/s
Initial angle (θ) = 22 degrees
Initial height (h₀) = 2.6 meters
To begin, we can split the initial velocity into its horizontal (v₀x) and vertical (v₀y) components. We'll use trigonometry to determine the magnitudes of these components.
v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)
Substituting the given values:
v₀x = 5.4 m/s * cos(22 degrees)
v₀y = 5.4 m/s * sin(22 degrees)
Next, let's determine how long the projectile will be in the air. We can use the vertical motion equation:
h = h₀ + v₀y * t - 1/2 * g * t²
Where:
h = final vertical position (which will be 0 since it lands on the ground)
h₀ = initial vertical position (2.6 meters)
v₀y = vertical component of initial velocity
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time of flight (unknown)
Since the projectile lands on the ground, h = 0. Rearranging the equation:
0 = 2.6 meters + (5.4 m/s * sin(22 degrees))t - 1/2 * (9.8 m/s²)t²
Now we can solve this quadratic equation for t using the quadratic formula. The equation will have two solutions, but we only want the positive one since time cannot be negative.
After obtaining the value of t, we can use it to calculate the horizontal displacement (range) the projectile travels:
Range = v₀x * t
Finally, we can determine the final velocity (v) and the final angle (θ') of the projectile when it hits the ground. Since acceleration solely affects the vertical component of velocity, the horizontal component remains constant.
v = √(v₀x² + v₀y²)
θ' = tan⁻¹(v₀y / v₀x)
By following these steps, you will be able to determine where the projectile will land, its final speed, and its angle of impact.