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Answer:
a) The height of the building is 12 metres.
b) The ball will take 6 seconds to hit the ground.
c) The maximum height of the ball is 16 metres and occurs 2 seconds after launch.
d) The ball have a height of 7 meters above the ground 5 seconds after launch.
Step-by-step explanation:
a) The roof of the building is represented by the initial height of the ball according to the function. If we know that [tex]t = 0\,s[/tex], the height of the building, measured in metres, is:
[tex]y = -(0\,s)^{2}+4\cdot (0\,s) + 12[/tex]
[tex]y = 12\,m[/tex]
The height of the building is 12 metres.
b) Let equalise the given polynomial and solve for [tex]t[/tex] to determine the time taken for the ball to hit the ground:
[tex]-t^{2}+4\cdot t +12 = 0[/tex] (1)
By the Quadratic Formula, we find the following solutions:
[tex]t_{1} = 6\,s[/tex] and [tex]t_{2} = -2\,s[/tex]
Since time is a positive variable, then the only solution that is physically reasonable is:
[tex]t = 6\,s[/tex]
The ball will take 6 seconds to hit the ground.
c) The maximum height of the ball occurs when speed is equal to zero. First, we differentiate the function and equalise to zero:
[tex]-2\cdot t + 4 = 0[/tex] (2)
[tex]t = 2\,s[/tex]
Lastly, we evaluate the function at given time:
[tex]y = -(2\,s)^{2}+4\cdot (2\,s)+12[/tex]
[tex]y = 16\,m[/tex]
The maximum height of the ball is 16 metres and occurs 2 seconds after launch.
d) We equalise the height formula to seven and solve the resulting polynomial:
[tex]-t^{2}+4\cdot t + 12 = 7[/tex]
[tex]-t^{2}+4\cdot t +5 = 0[/tex] (3)
By the Quadratic Formula, we get the following solutions:
[tex]t_{1} \approx 5\,s[/tex] and [tex]t_{2} \approx -1\,s[/tex]
The only solution that is physically reasonable is [tex]t = 5\,s[/tex].
The ball have a height of 7 meters above the ground 5 seconds after launch.