Motion in a Straight Line Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Class 12th Physics

A ball is thrown vertically upwards with a velocity of 19.6 ms^-1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m .$ The value of k is_______(use g = 9.8m/s²)

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Vishal Baghel | Contributor-Level 10

6 months ago

h = 39.2 + 19.6 = 58.8 m

Height above tower

$h' = \frac{u^{2}}{2 g} = \frac{19.6 * 19.6}{2 * 9.8}$

= 19.6

$A s, \frac{k}{5} = 78.4 \Rightarrow k = 392$

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A ball is thrown vertically upwards with a velocity of 19.6 ms-1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be (k5)m. The value of k is_______(use g = 9.8m/s2)

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A ball is thrown vertically upwards with a velocity of 19.6 ms^-1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m .$ The value of k is_______(use g = 9.8m/s²)