Question

A particle is thrown vertically upwards with a speed
of 10 m/s from top of a tower of height
20 m. One second after the projection, wind starts
blowing horizontally and imparts a constant
horizontal acceleration 10 m/s2 (along with gravity)
to the particle. The distance from the foot of the
tower, where the particle strikes the ground is
(g = 10 m/s2)​

Answer 1

Answer:

25 m

Explanation:

Time taken will remain same for both the motion(horizontal and vertical).

 For vertical motion: it is thrown upwards with the velocity of 10 m/s. At the highest point, velocity is 0

⇒ v = u + at     ⇒ 0 = 10 + (-10)t

⇒ t = 1 sec,    S = 5 cm  

t = 1 is that point from which the horizontal acceleration starts.

  From the highest point: where horizontal acceleration started: horizontal velocity is 0.

⇒ S = ut + 1/2 at²

⇒ 20 + 5 = (0)t + 1/2 (10)t²

⇒ 25 = 5t²

⇒ √5 = t

   

  In horizontal direction(from the highest point):

⇒ S = ut + 1/2 at²

⇒ S = (0)t + 1/2 (10)(√5)²

⇒ S = 25 m