stone is thrown with a velocity 20 m/s horizontally from top of a tower of height 19.6m. The horizontal range of the projectile i?

Take g = 9.8 m/s²

Consider the motion of the vertical component (uniform acceleration motion):
(downward to be positive):
Initial velocity, u(y) = 0 m/s
Displacement, s(y) = 19.6 m
Acceleration, a(y) = 9.8 m/s

s(y) = u(s) t + (1/2) a(y) t²
19.6 = 0 + (1/2) × 9,8 × t²
t = √(19.6 × 2 / 9.8)
Time taken, t = 2 s

Consider the motion of the horizontal component (uniform velocity motion):
s(x) = v(x) t
s(x) = 20 × 2
Horizontal range, s(x) = 40 m

let the stome take t sec to fall 19.6m, by s = ut + 1/2gt^2
=>19.6 = 0 + 1/2 x 9.8 x t^2
=>t = √4
=>t = 2sec
Thus horizontal range of the projectile (R), by s = ut
=>R = 20 x 2 = 40m