A camogie goalkeeper, on a level pitch, hit a ball straight up into the air - Leaving Cert Mathematics - Question 7 - 2019
Question 7
A camogie goalkeeper, on a level pitch, hit a ball straight up into the air.
The path that the ball travelled can be modelled by the function:
$$f(t) = -4t^2 + 16t... show full transcript
Worked Solution & Example Answer:A camogie goalkeeper, on a level pitch, hit a ball straight up into the air - Leaving Cert Mathematics - Question 7 - 2019
Step 1
At what height was the ball when it was hit by the goalkeeper?
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Answer
To find the height of the ball when it was hit by the goalkeeper, we need to evaluate the function at t=0:
f(0)=−4(0)2+16(0)+1=1.
Thus, the height of the ball when it was hit by the goalkeeper was 1 meter.
Step 2
Complete the table below to show the height of the ball at various intervals during the first 4 seconds of its flight.
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Answer
For the given time intervals, we calculate the height:
For t=0: f(0)=1 m.
For t=0.5: f(0.5)=13 m.
For t=1: f(1)=16 m.
For t=1.5: f(1.5)=17 m.
For t=2: f(2)=16 m.
For t=2.5: f(2.5)=13 m.
For t=3: f(3)=8 m.
For t=3.5: f(3.5)=1 m.
For t=4: f(4)=−4 m (not applicable, as it is on the ground).
The height values filled in the table would be:
Time (t)
Height (m)
0
1
0.5
13
1
16
1.5
17
2
16
2.5
13
3
8
3.5
1
4
0
Step 3
the length of time the ball was in the air from the time it was hit until it landed on the ground.
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Answer
From the graph, we can observe that the ball hits the ground at t=4 seconds. Therefore, the length of time the ball was in the air is:
4seconds.
Step 4
the length of time the ball was 10 m, or more, above the ground.
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Answer
By analyzing the table or graph, we see that the ball was above 10 meters between approximately t=0.5 seconds and t=2 seconds.
This gives:
2−0.5=1.5seconds.
Step 5
Find $f'(t)$, the derivative of $f(t) = -4t^2 + 16t + 1$.
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Answer
To find the derivative of the function, we take:
f′(t)=−8t+16.
Step 6
Use your answer from part (d)(i) to find the speed of the ball when it had been in the air for 4 seconds.
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Answer
Now, substituting t=4 into the derivative:
f′(4)=−8(4)+16=−32+16=−16m/s.
The speed of the ball when it had been in the air for 4 seconds is:
16m/s.
Step 7
Use your answer from part (d)(i) to find the value of $t$ for which the ball was descending and travelling at a speed of 8 metres per second.
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Answer
Setting the speed equal to -8 m/s (as it is descending):
−8t+16=−8.
Solving for t gives:
−8t=−8−16=−24t = rac{24}{8} = 3.
Thus, the value of t when the ball was descending at that speed is:
3seconds.
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