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anonymous

  • 4 years ago

A cannonball is fired into the air. It's height, h, in metres after t seconds is given by the relation: h=-4.6 (t-7.5)^2+285 a) what is the maximum height reached by the cannon and what time is this reached. b) what was the initial height of the cannon? c) what is the height of the cannon after 10.5 seconds? d) how long is the cannon in the air?

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  1. mathmate
    • 4 years ago
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    Hints: (a) The vertex form of a quadratic equation in t is y(t)=a(t-h)^2+k where (h,k) is the location of the vertex. (b) "initial" means at time t=0. Calculate h for t=0. (c) set t=10.5 and calculate h. (d)The time in the air is twice the time it takes to reach the vertex (highest point), so it would be 2h.

  2. anonymous
    • 4 years ago
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    i need help with the ans please.

  3. mathmate
    • 4 years ago
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    By comparing the formula given in (a), can you find h and k?

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