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A ball is thrown vertically upwards with a velocity of $19.6 \mathrm{~ms}^{-1}$ from the top of a tower. The ball strikes the ground after $6 \mathrm{~s}$ . The height from the ground up to which the ball can rise will be $\left(\frac{k}{5}\right) \mathrm{m}$ . The value of $\mathrm{k}$ is___________ (use $\mathrm{g=9.8 \mathrm{~m} / \mathrm{s}^{2}}$ )
Option: 1
392

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{h\rightarrow }$ height of building

$\mathrm{H\rightarrow }$ height from the ground upto which the ball can rise

From A to B

$\mathrm{v^2=u^2+2 a s}$

$\mathrm{0=(19.6)^2+2(-9.8)(H-h)}$

$\mathrm{H}-\mathrm{h}=19.6 \mathrm{~m} \rightarrow \text { (1) }\\$

From A to C

$\mathrm{s=u t+\frac{1}{2} a t^2}$

$\mathrm{(-h)=(+19.6) \times 6+\frac{1}{2}(-9.8)(6)^2}$

$\mathrm{-h=-9.8 \times 6}$

$\mathrm{h=58.8 \mathrm{~m} \rightarrow 2}$

$\mathrm{H=h+19.6}$

$\mathrm{=58.8+19.6}$

$\mathrm{H=78.4=\frac{k}{5}}$

$\mathrm{ K=392}$

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