Q124 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Physics

A ball is thrown vertically upwards with an initial velcoity $u$ reaches maximum height in $5 \mathrm{~s}$. The ratio of distance travelled by the ball in the 2 nd and 7 th second is (assume, $g=10$ $\mathrm{m} / \mathrm{s}^2$ )

  • A
    $8: 19$
  • B
    $16: 29$
  • C
    $16: 49$
  • Correct answer
    D
    None of these

Solution

When ball is at maximum height, its velocity is zero. From first equation of motion, $v=u-g t \Rightarrow 0=u-10 \times 5 \Rightarrow u=50 \mathrm{~m} / \mathrm{s}$ Distance travelled in $n$th second, $s_n=u+\frac{a}{2}(2 n-1)$ Distance travelled in 2nd second is given as $s_2=50-\frac{10}{2}(4-1)=50-15=35 \mathrm{~m}$ Distance travelled in 7th second is equal to distance travelled in 2nd second from highest point to vertically downward direction. $s_7=s_2{ }^{\prime}=0+\frac{10}{2}(2 \times 2-1)=15 \mathrm{~m}$ $\therefore \quad \frac{s_2}{s_7}=\frac{s_2}{s_2{ }^{\prime}}=\frac{35}{15}=\frac{7}{3}$ or $7: 23$

Hence, the correct answer is option: 'D'.

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