An object in launched directly upward at 54 feet per second (ft/s) from a platform 75 feet high. What will be the object's maximum height? When will it attain this height?
PE at top=PE at launch+Initial KE
I recommend you change feet to meters...
mgh=mg(distancehigh)+1/2 m vi^2 solve for h.
To find the maximum height of the object, we need to determine the time it takes for the object to reach its highest point.
We can use the formula for the time of flight of an object in projectile motion:
t = (Vf - Vi) / g
Where:
- t is the time of flight
- Vf is the final velocity (which is 0 at the highest point)
- Vi is the initial velocity
- g is the acceleration due to gravity (approximately 32.17 ft/s^2)
In this case, the initial velocity (Vi) is 54 ft/s and the acceleration due to gravity (g) is -32.17 ft/s^2 (negative because it acts against the direction of initial velocity).
Using these values, we can calculate the time of flight:
t = (0 - 54) / -32.17
t ≈ 1.680 seconds
Since the object reaches its maximum height halfway through the total flight time, we can find the maximum height by calculating the displacement at this time.
The displacement at any given time (t) is given by the equation:
h = Vi * t + (1/2) * g * t^2
Substituting the known values:
h = 54 * 1.680 + (1/2) * -32.17 * (1.680)^2
h ≈ 90.54 feet
Therefore, the object's maximum height is approximately 90.54 feet, and it will attain this height in about 1.680 seconds.