Question 111: Motion in 1D

Question

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 $$\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$$)

Accepted Answer

v = 19.6 m/s
t = 6s
Time taken in upward motion above tower = 2s
$$\Rightarrow$$ Time taken from top most point to ground = 4s
$$ \Rightarrow \sqrt {{{2h} \over g}}  = 4$$
$$h = {{16 	imes 9.8} \over 2} = 8 	imes 9.8$$
$$ \Rightarrow k = 8 	imes 9.8 	imes 5 = 392$$

Motion in 1D

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