With waht speed must a baseball be thrown upward from the ground in order to reach the top of the Washington Monment? (The top of the monument is 55o ft)
To determine the speed at which a baseball must be thrown upward from the ground to reach the top of the Washington Monument, we can use the principles of projectile motion.
First, we need to consider the vertical motion of the baseball. The initial vertical velocity (v₀y) of the baseball at the moment it is thrown upward is the speed we need to find. The final vertical position (y) we want to reach is the height of the Washington Monument, which is 550 ft (167.6 meters). The vertical acceleration (a) due to gravity is approximately -9.8 m/s² (assuming upward as positive).
We can use the following equation to find the time it takes for the baseball to reach the top of the Washington Monument:
y = v₀y * t + (1/2) * a * t²
Since the baseball starts from the ground (y₀ = 0), the equation simplifies to:
550 ft = (1/2) * (-9.8 m/s²) * t²
To solve for time (t), we need to convert the height to meters:
550 ft = (167.6 meters) = (1/2) * (-9.8 m/s²) * t²
Simplifying the equation further:
t² = (2 * 167.6 meters) / (-9.8 m/s²)
t² ≈ 34.0816
Taking the square root of both sides of the equation to solve for t:
t ≈ √34.0816 ≈ 5.83 seconds
Now that we know the time it takes for the baseball to reach the top of the Washington Monument, we can find the initial vertical velocity (v₀y):
v₀y = a * t
v₀y = (-9.8 m/s²) * 5.83 s
v₀y ≈ -57.23 m/s
Since upward is considered positive, we take the magnitude of the velocity:
v₀y ≈ 57.23 m/s
Therefore, the baseball must be thrown upward from the ground at a speed of approximately 57.23 m/s to reach the top of the Washington Monument.