Find the maximum height of a ball thrown upward from the top of a 640 ft tall building with an initial velocity of 128 ft/s
896
y = 640 + 128t - 16t^2
y' = 128 - 32t
y'=0 when t=4
so, what's y(4)?
To find the maximum height of a ball thrown upward, you can use the kinematic equation for vertical motion. The equation is:
h = (v^2 - u^2)/(2g)
Where:
h is the maximum height
v is the final velocity (0 m/s when the ball reaches its peak)
u is the initial velocity (128 ft/s)
g is the acceleration due to gravity (32.2 ft/s^2)
Substituting the given values into the equation:
h = (0 - 128^2)/(2 * 32.2)
Simplifying:
h = (-128^2)/64.4
h = -16384/64.4
h ≈ -254.37 ft
The negative sign indicates that the height is measured below the starting point. However, since the maximum height cannot be negative, we take the absolute value:
Maximum height ≈ | -254.37 | ft
≈ 254.37 ft
Therefore, the maximum height of the ball thrown upward from the top of a 640 ft tall building with an initial velocity of 128 ft/s is approximately 254.37 ft.
To find the maximum height of the ball, we can use the kinematic equation for vertical motion:
h = (v^2 - u^2) / (2g)
where:
h is the maximum height,
v is the final velocity (0 for a ball at maximum height),
u is the initial velocity,
g is the acceleration due to gravity.
Given:
u = 128 ft/s
g = 32 ft/s^2 (approximate value on Earth)
Substituting the known values into the equation:
h = (0^2 - 128^2) / (2 * 32)
Simplifying the equation:
h = (-16384) / 64
h = -256 ft
The negative sign indicates that the ball's maximum height is below the initial position. Therefore, the ball does not reach a maximum height and falls back to the ground before reaching the top of the building.