Answer:
The ball is dropped from a height of 36 feet
The bee and the ball will collide after approximately 1.3 seconds
The bee and the ball will collide at approximately 8 feet above the ground
The ball hits the ground after 1.5 seconds
Step-by-step explanation:
The complete question is
A ball dropped from the top of the building can be modeled by the function f(t)=-16t^2 + 36 , where t represents time in seconds after the ball was dropped. A bee's flight can be modeled by the function, g(t)=3t+4, where t represents time in seconds after the bee starts the flight.
The graph represents the situation.
select all that apply
1) The bee launches into flight from the ground.
2) The ball is dropped from a height of 36 feet.
3) The bee and the ball will collide after approximately 1.3 seconds.
4) The bee and the ball will collide after approximately 8 seconds.
5) The bee and the ball will collide at approximately 8 feet above the ground.
6) The bee and the ball will not collide.
7) The ball hits the ground after 1.5 seconds
see the attached figure to better understand the problem
Verify all the options
case 1) The bee launches into flight from the ground.
The statement is false
Because
we have
[tex]g(t)=3t+4[/tex]
For t=0
[tex]g(t)=3(0)+4=4\ ft[/tex]
That means ----> The bee launches into flight from a height 4 feet above the ground
case 2) The ball is dropped from a height of 36 feet
The statement is true
Because
For t=0
[tex]f(t)=-16(0)^2+36=36\ ft[/tex]
case 3) The bee and the ball will collide after approximately 1.3 seconds
The statement is true
Because
Equate f(t) and g(t)
[tex]-16t^2+36=3t+4[/tex]
[tex]-16t^2-3t+32=0[/tex]
solve the quadratic equation by graphing
The solution is t=1.324 sec
see the attached figure N 2
case 4) The bee and the ball will collide after approximately 8 seconds
The statement is false
Because, the bee and the ball will collide after approximately 1.3 seconds (see case 3)
case 5) The bee and the ball will collide at approximately 8 feet above the ground
The statement is true
Because
For t=1.324 sec
substitute the value of t in f(t) or g(t)
[tex]g(t)=3(1.324)+4=7.97\ ft[/tex]
case 6) The bee and the ball will not collide
The statement is false
see case 3) and case 5)
case 7) The ball hits the ground after 1.5 seconds
The statement is true
Because
we know that
The ball hit the ground when the value of f(t) is equal to zero
so
For f(t)=0
[tex]-16t^2+36=0[/tex]
[tex]t^2=\frac{36}{16}[/tex]
[tex]t=\frac{6}{4}=1.5\ sec[/tex]
