A body thrown vertically upwards with a speed of 19.6 ms -1 from the top of a tower returns to the earth in 10 seconds. What will be its height attained?

Question

A body thrown vertically upwards with a speed of 19.6 ms -1 from the top of a tower returns to the earth in 10 seconds. What will be its height attained? (A) 304 m (B) 308 m (C) 310 m (D) 312 m

Tardigrade Admin · Accepted Answer

Body covers equal distance in ascending or descending

h = u t + \frac{1}{2} g t^{2} (u = - v e)

(g = + v e

because after passing highest position motion is under gravity)

h = - 19.6 \times 10 + \frac{1}{2} \times 10 \times 100 (g = + v e)

= - 196 + 5 \times 100

= 500 - 196

h = 304 m

Q. A body thrown vertically upwards with a speed of 19.6ms−1 from the top of a tower returns to the earth in 10 seconds. What will be its height attained?

304 m

308 m

310 m

312 m

Body covers equal distance in ascending or descending h=ut+21​gt2(u=−ve) (g=+ve because after passing highest position motion is under gravity) h=−19.6×10+21​×10×100(g=+ve) =−196+5×100 =500−196 h=304m

Q. A body thrown vertically upwards with a speed of $19.6 m s^{- 1}$ from the top of a tower returns to the earth in $10$ seconds. What will be its height attained?

Body covers equal distance in ascending or descending
$h = u t + \frac{1}{2} g t^{2} (u = - v e)$
$(g = + v e$ because after passing highest position motion is under gravity)
$h = - 19.6 \times 10 + \frac{1}{2} \times 10 \times 100 (g = + v e)$
$= - 196 + 5 \times 100$
$= 500 - 196$
$h = 304 m$