Question

A ball is thrown is caught by the thrower after 6 seconds calculate the velocity with which the ball was thrown up and the maximum height attained by the boys and the distance of the ball below the highest point after 2 second gravitation = 10 M per second square

Answer 1

(a) Consider,

upward gravity = -9.8 m/s2

Total time (to and fro)= 6 s, so, upward journey = 6/2 = 3 s

Initial velocity (u) = ?

Final velocity (v) = 0 m/s

From First equation of motion,

Final velocity = Initial velocity +gt

or, 0 = u + (-9.8 X3)

or, u = 29.4 m/s

So, The velocity at which ball is thrown in the upward direction is 29.4 m/s.

(b) the maximum height (hmax) is reached by a ball

Second equation of motion

s = ut + 1/2 gt2

here s = the maximum height (hmax)

or , hmax =(29.4 X 3) + 1/2 (-9.8)(3)2

 hmax = 44.1 m

(c) Position of the ball after 4s is

From Second equation of motion

s = ut + 1/2 gt2

or, s = (29.4X4) + 1/2 X (-9.8)(4)2

or, s = 117.6 - 78.4 = 39.2 m