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)
Well, we would have to ignore air resistance which would have a major effect here but (using english units because you used feet):
v = Vo - 32 t
v = 0 at top
Vo = 32 t
h = 0 + Vo t - 16 t^2
550 = 32 t^2 - 16 t^2 = 16 t^2
t^2 = 34.4
t = 5.86 seconds
Vo = 32 t = 187 feet/second
To find 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 basic principles of projectile motion.
First, let's assume there is no air resistance. In this case, the vertical motion of the baseball can be analyzed independently from the horizontal motion.
The key equation we can use is the free-fall equation for vertical motion:
h = ut + (1/2)gt^2
Where:
h = height (the height of the Washington Monument is 550 ft)
u = initial vertical velocity (the speed at which the baseball is thrown upward)
g = acceleration due to gravity (approximately 32.2 ft/s^2, considering this is an Imperial unit problem)
t = time
At the top of the path, the final vertical velocity (v) will be 0 since the baseball reaches its maximum height and then starts moving downward.
Using these conditions, we can calculate the time it takes for the baseball to reach the top of the Washington Monument:
0 = u - gt
Simplifying the equation by isolating t:
t = u / g
Now we can substitute this value for t in the free-fall equation to solve for u:
h = (u * (u / g)) + (1/2)g(u / g)^2
Simplifying further:
2hg = u^2
Taking the square root of both sides:
u = sqrt(2hg)
Now we can substitute the given values into the equation:
u = sqrt(2 * 32.2 ft/s^2 * 550 ft)
Calculating the result:
u ≈ sqrt(35240 ft^2/s^2) ≈ 187.84 ft/s
Therefore, the baseball must be thrown upward with a speed of approximately 187.84 ft/s to reach the top of the Washington Monument.