Acatapult launches a rocket at an angle of 53.0° above the horizontal with an initial speed of 100 m/s. the rocket engine immediately starts a burn, and for 3.00 s the rocket moves along its initial line of motion with an acceleration of 30.0 m/s2. then its engine fails, and the rocket proceeds to move in free fall. find (a) the maxi- mum altitude reached by the rocket, (b) its total time of flight, and (c) its horizontal range.

Explanation:

A catapult launches a rocket at an angle of 53.0° above the horrizontal,

with an initial speed of 100m/s

The rocket engine immediately starts a burn, and for 3.00 s;  the rocket moves allong its initial line of motion with an acceleration of 30.0 m/s²

Then its engine fails and the rocket proceeds to move in free fall.

(a) We are first to find the maximum altitude reached by the rocket:

Final velocity (v) = acceleration × time = 30m/s² × 3.00 s = 90 m/s

The initial velocity (u) is 100m/s

Average velocity = = = 95 m/s

Displacement = average velocity × time taken = 95m/s × 3.00 s = 285 m

(b) We are then to find total time of flight.

The initial velocity when the catapult is starting to fall is 90 m/s

The final velocity of the catapult = 0 m/s

Average velocity = = = 45 m/s

Time taken to fall = average velocity ÷ gravitational acceleration (9.8 m/s²) = 45 m/s ÷ 9.8 m/s² = 4.591836735 seconds = 4.59 s (to nearest two decimal places)

Total time taken = 3.00 s + 4.59 s = 7.59 s

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .