Question

A ball is dropped from a platform $19.6\,m$ high. Its position function is -

• A. $x = - 4.9t^2 + 19.6 (0 \le t \le 1)$
• B. $x = - 4.9t^2 + 19.6 (0 \le t \le 2)$
• C. $x = - 9.8t^2 + 19.6 (0 \le t \le 2)$
• D. $x = - 4.9t^2 - 19.6 (0 \le t \le 2)$

Answer 1

Answer: (B)

We have, $a=\frac{d^{2} x}{d t^{2}}=-9.8$
The initial conditions are $x(0)=19.6$ and $v(0)=0$
So, $v=\frac{d x}{d t}=-9.8 t+v(0)=-9.8 t$
$\therefore x=-4.9 t^{2}+x(0)=-4.9 t^{2}+19.6$
Now, the domain of the function is restricted since the ball hits the ground after a certain time. To find this time we set $x=0$ and solve for $t$.
$0=4.9 t^{2}+19.6 \Rightarrow t=2$