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 6 s. The height from the ground up to which the ball can rise will be (k/5) m. The value of k is __________. (use g = 9.8 m/s²) Numerical answer type. Work it out, then open the answer.

Answer: 392. Upward positive, origin at top of tower: -H = ut - 1/2gt² = 19.6(6) - 4.9(36) -H = 117.6 - 176.4 = -58.8 ⇒ H = 58.8 m Rise above the tower: u²/2g = (19.6²)/(19.6) = 19.6 m Maximum height above ground = 58.8 + 19.6 = 78.4 m = k/5 k = 392

A ball is thrown vertically upwards with a velocity of 19.6 ms⁻¹ from… | JEE Main 2022 (28 Jul, Shift 2)

Q26JEE Main 2022 · 28 JUL · SHIFT 2Medium

Motion Under Gravity

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

Numerical answer type. Work it out, then open the answer.

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Answer: 392

Upward positive, origin at top of tower: $-H = ut - \frac{1}{2}gt^2 = 19.6(6) - 4.9(36)$
$-H = 117.6 - 176.4 = -58.8 \Rightarrow H = 58.8\text{ m}$
Rise above the tower: $\frac{u^2}{2g} = \frac{19.6^2}{19.6} = 19.6\text{ m}$
Maximum height above ground $= 58.8 + 19.6 = 78.4\text{ m} = \frac{k}{5}$
$k = 392$

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