A brick is thrown upward from the top of a building at an angle of 35° to the horizontal and with an initial speed of 15 m/s. If the brick is in flight for 2.7 s, how tall is the building? (in m)
could someone help me please!!!
We just need to do the vertical part of the problem. I assume it hit the ground where H = 0
Vi = 15 sin 35 = 8.6 m/s up
H = Hi + Vi t - 4.9 t^2
0 = Hi + 8.6 (2.7) - 4.9 (2.7)^2
Hi = 12.5 meters
I would be happy to help! But before we get to the answer, let's take a moment to appreciate the brick's ambitious attempt to reach new heights. Talk about aiming for the sky! Now, let's calculate the height of that building.
First, let's break down the problem. We know that the brick was thrown upward at an angle of 35° to the horizontal with an initial speed of 15 m/s. We're also given that the brick is in the air for 2.7 seconds.
To solve this problem, we can use the following equation to find the vertical component of the brick's initial velocity (Vy):
Vy = V * sin(θ)
Where V is the initial velocity (15 m/s) and θ is the angle of projection (35°).
Now, let's find Vy:
Vy = 15 m/s * sin(35°)
The vertical component of the initial velocity is approximately 8.62 m/s.
Next, we can use the kinematic equation to find the height (h) of the building:
h = Vy * t - 0.5 * g * t^2
Where t is the time of flight (2.7 seconds) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values:
h = 8.62 m/s * 2.7 s - 0.5 * 9.8 m/s^2 * (2.7 s)^2
After simplifying the equation, we find that the height of the building is approximately 29.64 meters.
So, the building is about 29.64 meters tall. I hope this helps! Remember, even in the world of physics, humor is an essential part of life.
Sure! I can help you with that.
To solve this problem, we can break it down into two components: horizontal motion and vertical motion.
First, let's find the horizontal distance the brick travels during the 2.7 seconds. Since there is no horizontal acceleration, we can use the equation:
distance = velocity * time
Given that the initial velocity is 15 m/s and the time is 2.7 s, we can calculate the horizontal distance:
horizontal distance = 15 m/s * 2.7 s = 40.5 m
Next, let's find the vertical distance the brick travels during the 2.7 seconds. The vertical motion of the brick can be described by the equation:
displacement = initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial vertical velocity is given as 15 m/s * sin(35°), the time is 2.7 s, and the acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction to the motion). Plugging these values into the equation, we get:
vertical displacement = (15 m/s * sin(35°)) * 2.7 s + (1/2) * (-9.8 m/s^2) * (2.7 s)^2
= 20.25 m - 35.07 m
= -14.82 m
The negative sign indicates that the brick is moving downward. Therefore, the brick falls 14.82 meters during the 2.7 seconds.
Finally, let's find the height of the building. The height of the building is equal to the initial vertical distance from which the brick was thrown, plus the distance it falls during the 2.7 seconds.
Given that the initial height is unknown, let's call it h. Therefore, we can write the equation as:
building height = h + (-14.82 m)
Since the brick was thrown from the top of the building, the initial height h is equal to the final height of the building.
So, building height = -14.82 m (Negative sign indicates downward direction)
Therefore, the height of the building is approximately 14.82 meters.
To find the height of the building, we need to break down the motion of the brick into its vertical and horizontal components.
Step 1: Find the vertical component of the initial velocity.
The vertical component can be found by multiplying the initial speed (15 m/s) with the sine of the launch angle (35°).
Vertical component = Initial speed × sin(angle)
Vertical component = 15 m/s × sin(35°)
Vertical component ≈ 15 m/s × 0.574
Vertical component ≈ 8.61 m/s
Step 2: Find the time taken for the brick to reach its maximum height.
Since the brick is thrown upward, the time taken to reach the maximum height is half of the total flight time.
Time to reach max height = Flight time / 2
Time to reach max height = 2.7 s / 2
Time to reach max height = 1.35 s
Step 3: Find the vertical displacement during the time to reach the maximum height.
The vertical displacement can be found using the equation:
Vertical displacement = Vertical component × Time to reach max height - 0.5 × g × (Time to reach max height)²
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Vertical displacement = 8.61 m/s × 1.35 s - 0.5 × 9.8 m/s² × (1.35 s)²
Vertical displacement ≈ 11.61 m - 9.95 m
Vertical displacement ≈ 1.66 m
Step 4: Find the total vertical displacement.
The total vertical displacement can be found by doubling the vertical displacement during the time to reach the maximum height.
Total vertical displacement = 2 × Vertical displacement
Total vertical displacement ≈ 2 × 1.66 m
Total vertical displacement ≈ 3.32 m
Therefore, the height of the building is approximately 3.32 meters.