Ms. chen is getting a new tile floor for her workshop. Her workshop is 15ft long and 11ft wide.
1 answer:
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Answer:
a) 40 feet
b) 54 ft/min
c) 4 mins
Step-by-step explanation:
Solution:-
- Kesha models the height ( h ) of the baton from the ground level but thrown from a platform of height hi.
- The function h ( t ) is modeled to follow a quadratic - parabolic path mathematically expressed as:
h ( t ) = −16t² + 54t + 40
Which gives the height of the baton from ground at time t mins.
- The initial point is of the height of the platform which is at a height of ( hi ) from the ground level.
- So the initial condition is expressed by time = 0 mins, the height of the baton h ( t ) would be:
h ( 0 ) = hi = -16*(0)^2 + 54*0 + 40
h ( 0 ) = hi = 0 + 0 + 40 = 40 feet
Answer: The height of the platform hi is 40 feet.
- The speed ( v ) during the parabolic path of the baton also varies with time t.
- The function of speed ( v ) with respect to time ( t ) can be determined by taking the derivative of displacement of baton from ground with respect to time t mins.
v ( t ) = dh / dt
v ( t )= d ( −16t² + 54t + 40 ) / dt
v ( t )= -2*(16)*t + 54
v ( t )= -32t + 54
- The velocity with which Kesha threw the baton is represented by tim t = 0 mins.
Hence,
v ( 0 ) = vi = -32*( 0 ) + 54
v ( 0 ) = vi = 54 ft / min
Answer: Kesha threw te baton with an initial speed of vo = 54 ft/min
- The baton reaches is maximum height h_max and comes down when all the kinetic energy is converted to potential energy. The baton starts to come down and cross the platform height hi = 40 feet and hits the ground.
- The height of the ball at ground is zero. Hence,
h ( t ) = 0
0 = −16t² + 54t + 40
0 = -8t^2 + 27t + 20
- Use the quadratic formula to solve the quadratic equation:
Answer: The time taken for the baton to hit the ground is t = 4 mins
Answer:
Step-by-step explanation:
I am assuming that you mean (4,−9) and (8, -3).
Hope this helps.
This problem is better understood with a given figure. Assuming that the flight is in a perfect northwest direction such that the angle is 45°, therefore I believe I have the correct figure to simulate the situation (see attached).
Now we are asked to find for the value of the hypotenuse (flight speed) given the angle and the side opposite to the angle. In this case, we use the sin function:
sin θ = opposite side / hypotenuse
sin 45 = 68 miles per hr / flight
flight = 68 miles per hr / sin 45
<span>flight = 96.17 miles per hr</span>
