Topics: Orders of magnitude, Problem solving, Real number Pages: 6 (266 words) Published: August 19, 2014
Yr 10

LCR Maths
By Adonis Chigeza

Understanding and Fluency Tasks
Task A
1. y = 1.2ūĚĎ• + 2.57
2. Interpolation: y = -3.43
Extrapolation: y = -8.23

Task B
a) The equation for the path of the ball is h = -0.1t^2 + 0.9t + 1 (h = height, t = time) b) The vertical height of the ball after 2. seconds2.664m
c) The maximum height reached by the ball is 3.025m
d) The time of with the ball is at maximum height of 3.025 is 4.5 seconds e) The total time in which the ball was in the air is 10 seconds f) The two times in which the ball was 1 metre above ground is 0 and 9

Adonis Chigeza 10C
LCR Mathematics

Problem Solving and Reasoning Task

Equation: y = -1.2ūĚíô2 + 8.4ūĚíô
a. The bridge is 7 metres wides so therefore it will successfully span the river with 2 metres to spare.
b. If a yacht has a 15 metre mask it will be unable to pass safely under the bridge because the bridge only has a vertical height 14.7 metres.

Adonis Chigeza 10C
LCR Mathematics

2. Equation: v= -0.2h2 + 2.4h
a. The horizontal distance covered by the rocket when it reached its maximum height of 7.2 metres was 6 metres.
b. The maximum height reached by the rocket was 7.2 metres.
c. At the horizontal distance of 9 metres from the launch site, there is a 5.2 metre wall and at that vertical distance, the rocket has a vertical distance 5.4 metre. That is not taking to account the dimensions of the rocket, however the rocket cannot have a diameter larger than 20cm otherwise it may not theoretically clear the wall.

Adonis Chigeza 10C
LCR Mathematics

3. Equation: y = ūĚíô2
Possible values
Ôā∑ a>1
Ôā∑ 0
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