A cannon fires a projectile at 75 m/s at an angle of 56.7 above the horizontal. How high does the projectile get off the ground?
S=displacement
g= gravitational pull(9.81 m/s²)
*Formula used is derived from here:
--> S= Vf² - Vi²/2a
2as= Vf² - Vi²
Vf²= Vi² + 2as
Change "a" to "g" since trajectory motion ang proble m and "s" ng "h" so magiging
(Vf²= Vi² + 2gh)
Since h=? And initial velocity is Zero
0=(VSinθ)² + 2gh
Divide both sides by 2(-g), since the gravity is pulling downward therfore the sign of constant g will be -9.81m/s²
h= - V² Sin²θ / 2(-g)
h= -(75 m/s)² Sin ² (56.7) / 2(-9.81 m/s²)
h= -3929.478 m²/s² / -19.62 m/s²
*Cancel all the excess units "m" will be left
h= 200.279 meters(ans.)
To find the maximum height that the projectile reaches, we can use the following equations of projectile motion:
1. Vertical velocity: vy = v * sin(θ)
2. Time of flight: t = 2 * vy / g
3. Maximum height: h = (vy^2) / (2 * g)
Given:
Initial velocity (v) = 75 m/s
Launch angle (θ) = 56.7 degrees
Acceleration due to gravity (g) = 9.8 m/s²
Step 1: Calculate the vertical velocity (vy):
vy = v * sin(θ)
= 75 * sin(56.7)
≈ 63.874 m/s
Step 2: Calculate the time of flight (t):
t = 2 * vy / g
= 2 * 63.874 / 9.8
≈ 13.03 s
Step 3: Calculate the maximum height (h):
h = (vy^2) / (2 * g)
= (63.874^2) / (2 * 9.8)
≈ 206.14 m
Therefore, the projectile reaches a maximum height of approximately 206.14 meters off the ground.
To determine how high the projectile gets off the ground, we need to analyze the motion of the projectile and use equations of motion. Specifically, we can use the vertical component of its initial velocity and the acceleration due to gravity to find the maximum height.
First, let's break down the initial velocity into its vertical and horizontal components. The vertical component of the initial velocity can be found using trigonometry:
Vertical component (V₀y) = V₀ * sin(θ)
= 75 m/s * sin(56.7°)
≈ 63.985 m/s
Next, we can use this vertical component to find how long it takes for the projectile to reach its maximum height. Using the equation:
v = u + at
where:
v = final velocity (0 m/s at maximum height, because the projectile momentarily stops)
u = initial velocity (63.985 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s²)
t = time
Rearranging the equation, we have:
t = (v - u) / a
= (0 - 63.985 m/s) / (-9.8 m/s²)
≈ 6.532 seconds
Now that we have the time it takes to reach maximum height, we can use this to calculate the maximum height itself. We can use the equation:
s = ut + (1/2)at²
where:
s = displacement (maximum height)
u = initial velocity (63.985 m/s)
t = time (6.532 seconds)
a = acceleration (acceleration due to gravity, -9.8 m/s²)
Substituting the values into the equation:
s = (63.985 m/s * 6.532 s) + (1/2)(-9.8 m/s²)(6.532 s)²
≈ 215.23 m
Therefore, the projectile reaches a height of approximately 215.23 meters off the ground.