A projectile is launched straight up from ground level with an initial velocity of 256 ft/sec.
a) When will the projectile's height above ground be 768 ft?
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There are three equations that you use to solve these sorts of equations.
1. final velocity = initial velocity + (acceleration x time)
2. displacement = initial displacement + (initial velocity x time) + (1/2)(acceleration)(time ^2)
3. (final velocity)^2 = (initial velocity)^2 + 2(acceleration)(displacement)
So what you want to do is is find the time at which the object reaches the maximum height. So you use equation 1. Use the final velocity as equal to zero (because at the max height the velocity will momentarily be zero), initial velocity as 256 ft per second and acceleration as negative 32.174 ft/s2 (acceleration of gravity in ft per second squared). And then you solve for time.
And that is your answer.
To find the total time that it takes to go up and then fall back down you just multiply that value by two because it takes the same time to go up as to come down.
You can then use that value for the time to substitute into the other equations for the max height reached.