Astunt man drives a car at a speed of 20 m/s off a 28-m-high cliff. the road leading to the cliff is inclined upward at an angle of 20°. how far from the base of the cliff does the car land? what is the car's impact speed?

car land from base at distance 39.66 m

impact speed = 45.17 m/s

Explanation:

given data

speed v = 20 m/s

height h  = 28 m

angle θ = 20°

to find out

How far from the base of cliff does the car land and car impact speed

solution

we consider here x is horizontal direction and

and y is vertical direction

so we find time first for climb 28 m high

use equation of motion for climb height

- h = v×sinθ × t - 0.5× a×t²  ............1

here h is height, t is time , a is acceleration and v is speed

put here all value

-28 = 20×sin20 × t - 0.5× 9.8 ×t²

t = 4.89 seconds

so

distance = ut = 20×cos20 × t = 20×cos20 × 4.86

distance = 39.66 m

so car land from base at distance 39.66 m

and

for impact speed we calculate vx and vy

we use equation of motion

v = u - at   ..................2

at vertical direction

vy = 20×sin20 - 9.8 ( 4.89) = -41.08 m/s  

at horizontal direction

vx = 20×cos20

vx = 18.79 m/s

so impact speed =    .................3

put here value

impact speed =

impact speed = 45.17 m/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 .