At the beginning of the school year, there were three 4th grade classes. Mrs. Smith’s class had 19 children, Mr. Day’s class has
1 answer:
You might be interested in
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