Question

a stone is thrown vertically up with an initial velocity of 49 metre per second from the top of a tower and reaches ground after 12 seconds. Find the height of the tower​

Answer 1

As,

s=u×t+1/2×a ×t square

=49×12+1/2×9.8×12×12

= 588+7056

= 7644m

Where s is, the height of the tower.

a is,the acceleration .

u is.,the initial velocity.

Answer 2

sign convention:

up : +ve

down : -ve

u = 49 m/s

a = - 9.8 m/s^2

t = 12 s

s = ut + (1/2) a×t^2

= 49 × 12 - (9.8 × 12 ×12 / 2)

= 49 × 12 (1 - 1.2)

= 49 × 12 × (- 0.2)

= - 117.6

so, height of the tower = 117.6 m ..... ans