A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, and it falls straight down to the ground below.
A) At a time of 6 seconds after it was thrown, how far above the ground is it?
The acceleration due to gravity is 10 m/s^2

700 m

Explanation:

Using the equations of motion,

g = 10 m/s², H = 1000 m,

Initial velocity, u = 20 m/s,

final velocity, v = ?

Total Time of fall, t = ?

a) But at t = 6s, g = 10 m/s², y = The height the ball has fallen through.

y = ut + gt²/2

y = (20×6) + 10(6²)/2

y = 300 m

Height of the ball above the ground at t = 6s is (Height of the cliff - height the ball has fallen through) = 1000 - 300 = 700 m

700 m.

Explanation:

Using equations of motion,

Given:

t = 6s

g = 10 m/s²y

S = vi*t + (g*t²)/2

= (20*6) + (10*6²)/2

= 300 m

Height above the ground = Height of the cliff - height of the motion

= 1000 - 300

= 700 m

we need some answer choices, or something like that

explanation: