A) \[x=-4.9{{t}^{2}}+19.6(0\le t\le 1)\]
B) \[x=-4.9{{t}^{2}}+19.6(0\le t\le 2)\]
C) \[x=-9.8{{t}^{2}}+19.6(0\le t\le 2)\]
D) \[x=-4.9{{t}^{2}}-19.6(0\le t\le 2)\]
Correct Answer: B
Solution :
[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, \[v=\frac{dx}{dt}=-9.8t+v(0)=-9.8t\] \[\therefore x=-4.9{{t}^{2}}+x(0)=-4.9{{t}^{2}}+19.6\] 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 \[t;0=-4.9{{t}^{2}}+19.6\Rightarrow t=2\]
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