A flower pot is thrown out of a window with a horizontal velocity of 8 m/s. If the window is 1.5 m off the ground, how far from the window does it land?
Provide me with an angle, because you ask inclination in a form of projectile.
h=gt²/2 =>t=sqrt(2h/g)
L=v(x)•t
A flower pot is thrown out of a window with a horizontal velocity of 8 m/s. If the window is 1.5 m off the ground, how far from the window does it land?
A. 0.3 m
B. 0.6 m
C. 2.4 m
D. 4.4 m
It's D.
To answer this question, we need to analyze the motion of the flower pot both vertically and horizontally. Let's break it down step by step:
1. First, let's consider the vertical motion of the flower pot. Since the flower pot is thrown horizontally, there is no initial vertical velocity. Thus, we can use the equation for vertical displacement:
Δy = v₀y * t + 0.5 * a * t²
Where:
- Δy is the vertical displacement (which is 1.5 m because the window is 1.5 m off the ground).
- v₀y is the initial vertical velocity (which is 0 m/s since the pot is thrown horizontally).
- t is the time it takes for the flower pot to land.
- a is the acceleration due to gravity (which is approximately 9.8 m/s²).
Since Δy = 1.5 m and v₀y = 0 m/s, the equation simplifies to:
1.5 = 0.5 * 9.8 * t²
Rearranging the equation gives us:
t² = 1.5 / (0.5 * 9.8)
t² = 0.30612245
Taking the square root of both sides, we find:
t ≈ 0.553 s
2. Now, let's consider the horizontal motion of the flower pot. The horizontal velocity is given as 8 m/s. Since there is no horizontal acceleration (assuming no air resistance), the horizontal displacement (x) can be calculated using the equation:
x = v₀x * t
Where:
- x is the horizontal displacement.
- v₀x is the initial horizontal velocity (which is 8 m/s).
- t is the time we found in step 1.
Plugging in the values, we have:
x = 8 * 0.553
x ≈ 4.424 m
Therefore, the flower pot will land approximately 4.424 meters away from the window.