Question

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. 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:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

$y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2$.

where y(t) represent the height from the ground. For our problem, the initial height will be:

$y_0 \ = \ 1000 m$.

The initial velocity:

$v_0 = - 20 \frac{m}{s}$,

take into consideration the minus sign, that appears cause the ball its thrown down.  The same minus appears for the acceleration:

$a=-10\frac{m}{s}$

So, the equation for our problem its:

$y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2$.

Taking t=6 s:

$y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2$.

$y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2$.

$y(6 \ s) = \ 1000 m \ - 120 m - 180 m$.

$y(6 \ s) = \ 1000 m \ - 300 m$.

$y(6 \ s) = \ 700 m$.

So this its the height of the ball 6 seconds after being thrown.