If a projectile has a launching angle of 52 degrees about the horizontal and an initial speed of 18 m/s, what is the highest barrier that the projectile can clear?
Vo = 18m/s @ 52 deg.
Vo(hor) = 18cos52 = 11.08m/s.
Vo(ver) = 18sin52 = 14.18m/2,up.
Vf^2 = Vo^2 + 2gd = 0 @ max height.
(14.18)^2 + 2(-9.8)d = 0,
201.07 -19.6d = 0,
d = 10.3m. = max barrier that can be
cleared.
Well, if I were a projectile trying to clear a barrier, I would definitely need some good leaping skills! Now, let's calculate this together.
Given that the launching angle is 52 degrees, we know this is related to the maximum height of the projectile. To determine the highest barrier it can clear, we need to find the maximum height reached by the projectile.
Now, the maximum height can be calculated using the formula:
H = (V² * sin²θ) / (2 * g)
where:
H is the maximum height,
V is the initial velocity (18 m/s),
θ is the launching angle (52 degrees),
and g is the acceleration due to gravity (approximately 9.8 m/s²).
Plug in the values into the formula, and we can compute the highest barrier the projectile can clear. Just keep in mind that my jokes might not be able to clear any barriers because laughter doesn't give me the ability to fly! So, let's calculate it!
To find the height of the highest barrier that the projectile can clear, we need to analyze the projectile's motion. Here are the step-by-step calculations:
Step 1: Split the initial velocity into its horizontal and vertical components:
The horizontal component of velocity (Vx) remains constant throughout the motion. The vertical component of velocity (Vy) will change due to the effect of gravity.
Given:
Initial speed (Vi) = 18 m/s
Launch angle (θ) = 52 degrees
Vx = Vi * cos(θ)
Vy = Vi * sin(θ)
Vx = 18 * cos(52)
Vx = 18 * 0.6157
Vx ≈ 11.0814 m/s (rounded to 4 decimal places)
Vy = 18 * sin(52)
Vy = 18 * 0.7934
Vy ≈ 14.2812 m/s (rounded to 4 decimal places)
Step 2: Calculate the time of flight (T) of the projectile.
The time of flight is the total time it takes for the projectile to reach the same vertical position from where it was launched.
T = (2 * Vy) / g
where g = acceleration due to gravity = 9.8 m/s^2
T = (2 * 14.2812) / 9.8
T ≈ 2.9124 seconds (rounded to 4 decimal places)
Step 3: Calculate the maximum height (H) reached by the projectile.
The maximum height is the height reached by the projectile when its vertical velocity becomes zero (at the top of its trajectory).
H = (Vy^2) / (2 * g)
H = (14.2812^2) / (2 * 9.8)
H ≈ 10.3942 meters (rounded to 4 decimal places)
Therefore, the projectile can clear a barrier with a maximum height of approximately 10.3942 meters.
To find the highest barrier that the projectile can clear, we need to determine the maximum height it reaches during its flight. This can be calculated using the equation of motion for vertical motion.
In this case, the projectile is launched at an angle of 52 degrees, so we need to find the vertical component of the initial velocity (v0y) and the time it takes for the projectile to reach its maximum height (t).
First, let's find the vertical component of the initial velocity (v0y). This can be calculated using the equation:
v0y = v0 * sin(theta)
Where:
- v0 is the initial speed of the projectile, which is 18 m/s
- theta is the launching angle, which is 52 degrees
Using this equation, we can calculate v0y as:
v0y = 18 m/s * sin(52 degrees)
v0y ≈ 14.24 m/s
Next, we can calculate the time it takes for the projectile to reach its maximum height using the equation of motion:
t = v0y / g
Where:
- g is the acceleration due to gravity, which is approximately 9.8 m/s^2
Plugging in the values, we have:
t = 14.24 m/s / 9.8 m/s^2
t ≈ 1.45 s
Now that we know the time it takes to reach the maximum height, we can calculate the maximum height (h) using the equation:
h = v0y^2 / (2 * g)
Plugging in the values, we have:
h = (14.24 m/s)^2 / (2 * 9.8 m/s^2)
h ≈ 10.29 m
Therefore, the projectile can clear a barrier with a maximum height of approximately 10.29 meters.