a 275-kilogram object starts from rest and accelerates on level ground up to a speed of

41 meters per second over 49.3 meters. it continues at that speed for another 15

meters. at the end of the 15 meters, the object is launched forward at a 37° angle.

a. how much time does it take for the object to reach its final velocity?

b. how much force did the object apply in order to reach that final speed?

c. what are the maximum range and height of the object when it is launched?

d. at what angle would the object have to be launched to reach a height of 50

meters above the ground at the same initial velocity?

e. describe a realistic scenario in which this might happen.

Explanation:

a )

mass m = 275 kg , final velocity v = 41 m/s

distance,  s = 49.3 m

acceleration a = ?

v² = u² + 2as

41² = 2 a x 49.3

a = 17.05 m/s²

time t = v - u / a

t = 41 / 17.05

= 2.4 s

b )

force = mass x acceleration

= 275 x 17.05

= 4688.75 N

c ) maximum range = v² / g

= 41²/9.8

= 171.5 m

height ( maximum )

v² sin²45 / 2g ( at angle of projectile of 45 , range and height is maximum.)

= 41²/ 2 x 9.8

= 85.76 m

d )

Let the angle be θ

height h = v² sin²θ / 2g

50 = 41² sin²θ /2g

sin²θ  = .58

sinθ = .76

50°.

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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 .