JEE Main 2022 · 28 Jul · Shift 2 · Physics Q21

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PhysicsInteger+4 / -1Question 21 of 90

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 $\left(\dfrac k5\right)$ m. The value of k is ______ (use g = 9.8 m/s $^2$ )

Your answer

Answer and worked solution

Official answer: 392

Tower. $-h=19.6\times6-4.9\times36$ , so $h=58.8$ m.

Rise above the tower $\frac{19.6^2}{19.6}=19.6$ m, so the top is 78.4 m $=\frac{392}5$ m.

Answer: 392

Topics:Motion in a straight line

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JEE Main 2022 · 28 Jul · Shift 2, Physics question 21