We are given that an object follows the next parametric equations:

Where:

Now, we substitute the given values for the horizontal reach we get:

For part B we are asked to determine the time that the object would take to reach a horizontal distance of 400 feet. To do that we will substitute the value of "x = 400" in the equation:
![400=\operatorname{\lparen}128cos60)t]()
Now, we solve for the time "t" by dividing both sides by "128cos60":

Solving the operations:

Therefore, it would take 6.25 seconds for the object to reach the 400 ft.
Part C we are asked to determine the height when the object reaches the 400 ft. To do that we use the equation for the height "y":

Substituting the values we get:

Now, we substitute the value of time "t = 6.25s" which is the time it takes the object to reach the 400 ft:

Solving the operations:

Therefore, the height is 75.82 ft.