A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take g = 10 m/s2.

Sol. For last 6 m distance travelled s = 6 m, u = ? t = 0.2 sec, a = g = 9.8 m/s2 S = ut + ½ at2 ⇒ 6 = u (0.2) + 4.9 x 0.04 ⇒ u = 5.8/0.2 = 29 m/s. For distance x, u = 0, v = 29 m/s, a = g = 9.8 m/s2 S = (V^2-u^2)/2a = (〖29〗^2 - 0^2 )/(2 x 9.8 ) = 42.05 m Total distance = 42.05 + 6 = 48.05 = 48 m.

Sol: According to the question the ball is yet to cover the last 6 m above the ground and the ball takes 0.2 secs to cover the distance above the last 6 meter.So we need to find the distance above the last 6 m and then add 6 to it to get the final answer.So, let the total height be `h` and the covered distance be `h-6`.As the ball is dropped freely:-Initial velocity `u` = 0Distance `x` = h-6Time `t` = 0.2 secsAcceleration `a` = `g` = 10 m/s²Now according to kinematic formula:x = ut + 1/2gt²Substituting the values:-h-6 = (0)(0.2) + (1/2)(10)(0.04)h-6 = 0.2h = 0.2 + 6h = 6.2 m Therefore, the ball was dropped from the height of 6.2m.