A projectile is fired with an initial velocity of 300m/s.upward at an angle of 30 degree with horizontal from a point 80m above the level plain. What horizontal distance will it cover before it strikes the plain?
vertical problem first
h = Hi + Vi t - 4.9 t^2
0 = 80 + 300 (sin 30) t - 4.9 t^2
solve quadratic for t. Use positive t, the other root was on the way up underground.
then
Horizontal problem at constant u = 300 cos 30
d = u * t
To find the horizontal distance covered by the projectile before it strikes the plane, we can use the equations of projectile motion.
Step 1: Determine the time it takes for the projectile to reach the level plane.
We can use the vertical motion equation, which relates the vertical displacement, initial vertical velocity, time, and acceleration due to gravity:
Δy = v₀y * t + (1/2) * g * t²
Given:
Δy = 80m (vertical displacement)
v₀y = v₀ * sin(θ) = 300 * sin(30°) ≈ 150m/s (initial vertical velocity)
g = 9.8m/s² (acceleration due to gravity)
Plugging in the values:
80 = 150 * t + (1/2) * 9.8 * t²
This is a quadratic equation. Simplifying it, we get:
4.9 * t² + 150 * t - 80 = 0
Step 2: Solve the quadratic equation to find the time.
We can solve this equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
In this equation:
a = 4.9
b = 150
c = -80
Plugging in the values and simplifying, we get:
t = (-150 ± √(150² - 4 * 4.9 * (-80))) / (2 * 4.9)
Simplifying further:
t ≈ (-150 ± 332.45) / 9.8
Now we have two possible values for t:
t₁ ≈ (-150 + 332.45) / 9.8 ≈ 19.56s
t₂ ≈ (-150 - 332.45) / 9.8 ≈ -20.57s
Since time cannot be negative, we discard the negative value and take t = 19.56s.
Step 3: Find the horizontal distance covered by the projectile.
We can use the horizontal motion equation, which relates the horizontal displacement, initial horizontal velocity, and time:
Δx = v₀x * t
Given:
v₀x = v₀ * cos(θ) = 300 * cos(30°) ≈ 259.8m/s (initial horizontal velocity)
t ≈ 19.56s (time)
Plugging in the values:
Δx = 259.8 * 19.56 ≈ 5074.69m
Therefore, the horizontal distance covered by the projectile before it strikes the plane is approximately 5074.69 meters.
To find the horizontal distance covered by the projectile, we need to analyze its vertical and horizontal motion separately.
First, let's consider the vertical motion. We can use the following equation of motion to find the time it takes for the projectile to reach the ground:
y = u * sin(theta) * t - 0.5 * g * t^2
Where:
- y is the vertical displacement (80m in this case),
- u is the initial velocity (300 m/s),
- theta is the launch angle (30 degrees),
- t is the time taken,
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since we want to find the time it takes for the projectile to reach the ground, we set y to 0 and solve for t:
0 = 300 * sin(30) * t - 0.5 * 9.8 * t^2
This equation is a quadratic equation, which we can solve using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -0.5 * 9.8, b = 300 * sin(30), and c = 0. Plugging in these values, we can find the time it takes for the projectile to reach the ground.
Once we have the time, we can use the horizontal component of the initial velocity to calculate the horizontal distance. The horizontal velocity (v_x) can be found using:
v_x = u * cos(theta)
Then, the horizontal distance (d) is given by:
d = v_x * t
Using the values of u, theta, and the time obtained from the previous steps, we can calculate the horizontal distance covered by the projectile before it strikes the ground.