Question

A projectile is fired vertically and its height h, in meters, as a function of time t, in seconds, is given by the function h(t) = - 2t ^ 2 + 20t From this we ask: (a) How long did it take for the projectile to touch the ground? (b) What is the maximum height reached by the projectile?

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Answer to a math question A projectile is fired vertically and its height h, in meters, as a function of time t, in seconds, is given by the function h(t) = - 2t ^ 2 + 20t From this we ask: (a) How long did it take for the projectile to touch the ground? (b) What is the maximum height reached by the projectile?

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Birdie
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94 Answers
Para determinar quando o projétil irá tocar o solo, precisamos encontrar o valor de t quando h(t) = 0 . Portanto, temos:

h(t) = -2t^2 + 20t

Para determinar o tempo que o projétil leva para tocar o solo, resolvemos a equação:

-2t^2 + 20t = 0

Fatorando -2t , temos:

-2t(t - 10) = 0

Então, temos duas possibilidades:

1. -2t = 0 , que implica em t = 0
2. t - 10 = 0 , que implica em t = 10

O projétil leva 10 segundos para tocar o solo.

Para determinar a altura máxima atingida pelo projétil, vamos encontrar o vértice da parábola h(t) = -2t^2 + 20t .

O eixo de simetria de uma parábola é dado por t = -\frac{b}{2a} . No caso de h(t) = -2t^2 + 20t , onde a = -2 e b = 20 , temos:

t = -\frac{20}{2*(-2)} = -\frac{20}{-4} = 5

Portanto, o projétil atinge a altura máxima em t = 5 segundos.

Para calcular a altura máxima, substituímos t = 5 na equação h(t) :

h(5) = -2(5)^2 + 20(5) = -2*25 + 100 = -50 + 100 = 50

Assim, a altura máxima atingida pelo projétil é 50 metros.

\textbf{Respostas:}

(a) O projétil demorou 10 segundos para tocar o solo.

(b) A altura máxima atingida pelo projétil foi de 50 metros.

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