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 $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$)
Answer (integer)
392
Solution
<p>v = 19.6 m/s</p>
<p>t = 6s</p>
<p>Time taken in upward motion above tower = 2s</p>
<p>$\Rightarrow$ Time taken from top most point to ground = 4s</p>
<p>$\Rightarrow \sqrt {{{2h} \over g}} = 4$</p>
<p>$h = {{16 \times 9.8} \over 2} = 8 \times 9.8$</p>
<p>$\Rightarrow k = 8 \times 9.8 \times 5 = 392$</p>
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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