Question

A little boy is standing at the edge of a cliff 500 m high. He throws a ball straight upward at an initial speed of 20 m/s, and it falls straight to the ground below. 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/s2 .

Answer 1

Answer: At a time of 6 seconds after it was thrown, how far above the ground is it?

Step-by-step explanation:

We can start by using the kinematic equation for vertical motion:

y = y0 + v0t + (1/2)at^2

where y is the final height, y0 is the initial height, v0 is the initial velocity, t is the time, and a is the acceleration due to gravity.

We can use this equation to find the height of the ball at a time of 6 seconds after it was thrown. The initial height, y0, is 500 m, the initial velocity, v0, is 20 m/s (upward), the acceleration due to gravity, a, is -10 m/s^2 (downward), and the time, t, is 6 seconds.

Plugging in these values, we get:

y = 500 m + (20 m/s)(6 s) + (1/2)(-10 m/s^2)(6 s)^2

Simplifying, we get:

y = 500 m + 120 m - 180 m

y = 440 m

Therefore, at a time of 6 seconds after it was thrown, the ball is 440 meters above the ground.

{Hope This Helps! :)}