Series, help! (1 Viewer)

excellence1 · Dec 6, 2015

A ball thrown from the top of the tallest building the Buruj Khalifa in Dubai falls 3m the 1st sec, 9m the 2nd sec and 15 in the 3rd sec.
a) How far does the ball fall in the 15th sec
If the height of the Burij Khalifa is 828 m, find how long it would take for the ball to hit the ground.

InteGrand · Dec 6, 2015

excellence1 said:
A ball thrown from the top of the tallest building the Buruj Khalifa in Dubai falls 3m the 1st sec, 9m the 2nd sec and 15 in the 3rd sec.
a) How far does the ball fall in the 15th sec
If the height of the Burij Khalifa is 828 m, find how long it would take for the ball to hit the ground.

$(a) Let the distance fallen in the $n^{\text{th}}$ second be $u_n$ (in metres). So $u_1 = 3,u_2 = 9, u_3 = 15$, etc. We assume the distances fallen form an arithmetic progression. The starting term is $u_1 = 3$ and the common difference is $d=6$.$

$Part (a) asks us to find the value of $u_{15}$. This is $u_{15}=u_1 + 14 d = 3 + 14 \cdot 6 =89$. Hence the answer to part (a) is 89 m.$

$(b) Let the required no. of seconds be $N$, so that $u_1 + u_2 + \cdots + u_N = 828$. We need to find the value of $N$.$

$Using the sum of an A.P. formula, we have$

$\begin{align*} \frac{N}{2}\left(2u_1 + \left(N-1\right)d \right)&= 828 \\ \Rightarrow \frac{N}{2}\left(6 + 6 \left(N-1\right) \right)&= 828 \quad (\text{as }u_1=3,d=6) \\ \Rightarrow \frac{N}{2}\times 6N &= 828 \\ \Rightarrow 3N^2 &= 828 \\ \Rightarrow N^2 &= 276 \\ \Rightarrow N&= \sqrt{276} = 16.613....\end{align*}$

$So the ground is hit during the $17^{\text{th}}$ second.$

Series, help! (1 Viewer)

excellence1

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InteGrand

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