A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and with an initial speed of 15m/s. If the brick is in the flightfor 3.0 s, how tall is the building?
Vo = 15m/s[25o],
Yo = 15*sin25 = 6.3 m/s. = Ver. component of initial velocity.
Y = Yo + g*Tr = 0,
6.3 + (-9.8)Tr = 0,
Tr = 0.65 s. = Rise time.
Tr + Tf = 3 s,
0.65 + Tf = 3,
Tf = 2.35 s. = Fall time.
ho = 0.5g*Tf^2 = 4.9 * 2.35^2 = 27.1 m. above gnd.
Y^2 = Yo^2 + 2g*h = 0,
6.3^2 + (-19.6)h = 0,
h = 2.03 m. above the bldg.
hb + h = ho,
hb + 2.03 = 27.1,
hb = 25.1 m. = Ht. of the bldg.
https://www.jiskha.com/questions/682094/A-brick-is-thrown-upward-from-the-top-of-a-building-at-an-angle-of-25-to-the-horizontal
Why did the brick become an expert in high-rise buildings? Because it was aiming for the top! Let's calculate the height of the building now.
To solve this problem, we'll need to break down the motion of the brick into horizontal and vertical components.
First, let's find the vertical component of the initial velocity when the brick was thrown upwards. We can use the formula:
Vsinθ = 15 m/s * sin(25°)
Vsinθ ≈ 6.42 m/s
Now, we'll use the vertical component to calculate the time it takes for the brick to reach its maximum height. We'll use the formula:
Vy = Voy - gt
0 = 6.42 m/s - 9.8 m/s^2 * tmax
Solving for tmax gives us:
tmax ≈ 0.656 seconds
Since the total flight time is given as 3.0 seconds, we know that the time taken for the brick to reach its maximum height is half the total flight time. Therefore, the brick will take another 2.344 seconds to come back down.
Now we'll find the height of the building using the formula:
h = Voy * t - 0.5 * g * t^2
h = 6.42 m/s * 2.344 s - 0.5 * 9.8 m/s^2 * (2.344 s)^2
h ≈ 17.8 meters
So, the height of the building is approximately 17.8 meters. Let's hope the brick didn't make a dent!
To find the height of the building, we need to determine the vertical displacement of the brick during its flight. We can use the following equations of motion:
1. Vertical displacement equation:
Δy = v0y * t + (1/2) * a * t^2
2. Initial vertical velocity equation:
v0y = v0 * sin(θ)
3. Acceleration due to gravity equation:
a = -g
where:
- Δy is the vertical displacement
- v0y is the initial vertical velocity
- t is the time of flight
- v0 is the initial speed of the brick
- θ is the launch angle
- g is the acceleration due to gravity (approximately -9.8 m/s^2)
Let's substitute the given values into the equations:
v0y = v0 * sin(θ) = 15 m/s * sin(25°) ≈ 6.32 m/s
a = -g = -9.8 m/s^2
t = 3.0 s
Using equation 1, we can solve for Δy:
Δy = v0y * t + (1/2) * a * t^2
= 6.32 m/s * 3.0 s + (1/2) * -9.8 m/s^2 * (3.0 s)^2
= 18.96 m + (-44.1 m)
≈ -25.14 m
The negative sign indicates that the displacement is in the opposite direction to the positive vertical direction, meaning the brick falls downward.
Since we're interested in the height of the building, we need to take the absolute value of the displacement:
height = |Δy|
= |-25.14 m|
= 25.14 m
Therefore, the height of the building is approximately 25.14 meters.
To solve this problem, we need to use the equations of motion for projectile motion. The brick is thrown upward at an angle of 25° to the horizontal with an initial speed of 15 m/s. We are given that the time of flight is 3.0 seconds.
First, let's break down the motion into horizontal and vertical components:
Horizontal component:
The initial horizontal velocity (Vx) remains constant throughout the flight because there is no acceleration in the horizontal direction. Therefore, Vx = 15 m/s.
Vertical component:
The initial vertical velocity (Vy) can be found by using the equation Vy = V * sin(θ), where V is the initial speed and θ is the angle of projection.
Vy = 15 m/s * sin(25°) ≈ 6.42 m/s.
Using the equation of motion for the vertical motion, we can find the height of the building. The equation is:
h = Vy * t - 0.5 * g * t^2
Where h is the height, Vy is the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity which is approximately 9.8 m/s^2.
Plugging in the values, we get:
h = 6.42 m/s * 3.0 s - 0.5 * 9.8 m/s^2 * (3.0 s)^2
h = 19.26 m - 44.1 m
h ≈ -24.84 m
Since the height cannot be negative, we take the absolute value to get the final height of the building:
| h | ≈ |-24.84 m| = 24.84 m
Therefore, the height of the building is approximately 24.84 meters.