IB DP Physics 2.1 – Motion: IB Style Question Bank HL Paper 1

Ans: B

Assuming constant vertical acceleration, no horizontal acceleration, and no air resistance:

\(y(t) =y_0+\dot{y_0}t+\frac{1}{2}\ddot{y}t^2\)

Then when \(y(t_1)=y_1\)

\(\frac{\ddot{y}}{2}t_1^2+\dot{y_0t_1}+(y_0-y_1)=0)\)

so by quadratic formula

\(t_1=\frac{-\dot{y_0}\pm \sqrt{\dot{y}_0^2-4 \times \frac{\ddot{y}}{2}\times(y_0-y_1)}}{2 \times \frac{\ddot{y}}{2}}\)

Given

\(\dot{y_0}=0\;m/s\)
\(\ddot{y}=g< 0\)
\(t_1\geq 0\)

we can reduce that to

\(t_1=\frac{\sqrt{-2g(y_0-y_1)}}{-g}=\sqrt{\frac{2(y_1-y_0)}{g}}\)

In that same time the stone has moved \(\dot{x_0}t_1 \;m\) horizontally.

Given

\(\dot{x_0}=15 ms^{-1}\)
\(y_0=80m \)
\(y_1=0 m\)
\(g=-9.81 ms^{-2}\)

we get

\(x_1\) =  60 m