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 / s 2 )

Question

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 / s 2 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

t a =( u / g )=(19.6/9.8)=2 s t d =6-2 s =√(2 h max / g ) ⇒ h max =(16 × 9.8/2)=(392/5)

Q. A ball is thrown vertically upwards with a velocity of $19.6 m s^{- 1}$ 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 $(\frac{k}{5}) m$ . The value of $k$ is ..... (use $g \equiv 9.8 m / s^{2}$ )

$t_{a} = \frac{u}{g} = \frac{19.6}{9.8} = 2 s$
$t_{d} = 6 - 2 s = \frac{2 h _{m a x}}{g}$
$\Rightarrow h_{m a x} = \frac{16 \times 9.8}{2} = \frac{392}{5}$

Q. A ball is thrown vertically upwards with a velocity of 19.6ms−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 (5k​)m. The value of k is ..... (use g≡9.8m/s2 )

ta​=gu​=9.819.6​=2s td​=6−2s=g2hmax​​​ ⇒hmax​=216×9.8​=5392​

Q. A ball is thrown vertically upwards with a velocity of $19.6 m s^{- 1}$ 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 $(\frac{k}{5}) m$ . The value of $k$ is ..... (use $g \equiv 9.8 m / s^{2}$ )

$t_{a} = \frac{u}{g} = \frac{19.6}{9.8} = 2 s$
$t_{d} = 6 - 2 s = \frac{2 h _{m a x}}{g}$
$\Rightarrow h_{m a x} = \frac{16 \times 9.8}{2} = \frac{392}{5}$