Question
A ball is dropped from a platform $$19.6\,m$$ high. Its position function is :
A.
$$x = - 4.9{t^2} + 19.6\left( {0 \leqslant t \leqslant 1} \right)$$
B.
$$x = - 4.9{t^2} + 19.6\left( {0 \leqslant t \leqslant 2} \right)$$
C.
$$x = - 9.8{t^2} + 19.6\left( {0 \leqslant t \leqslant 2} \right)$$
D.
$$x = - 4.9{t^2} - 19.6\left( {0 \leqslant t \leqslant 2} \right)$$
Answer :
$$x = - 4.9{t^2} + 19.6\left( {0 \leqslant t \leqslant 2} \right)$$
Solution :
We have, $$a = \frac{{{d^2}x}}{{d{t^2}}} = - 9.8$$
The initial conditions are $$x\left( 0 \right) = 19.6{\text{ and }}v\left( 0 \right) = 0$$
$$\eqalign{
& {\text{So, }}v = \frac{{dx}}{{dt}} = - 9.8t + v\left( 0 \right) = - 9.8t \cr
& \therefore \,x = - 4.9{t^2} + x\left( 0 \right) = - 4.9{t^2} + 19.6 \cr} $$
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 $$t;\,\,0 = - 4.9{t^2} + 19.6 \Rightarrow t = 2$$