Answer
(a) $0.208m$
(b) $0.0520m$
Work Step by Step
(a) We can find the required vertical height as follows:
$K.E_B+U_B=K.E_A+U_A$
$\implies \frac{1}{2}mv_B^2+0=0+mgy_A$
This simplifies to:
$y_A=\frac{v_B^2}{2g}$
We plug in the known values to obtain:
$y_A=\frac{(2.02)^2}{2(9.81)}$
$y_A=0.208m$
(b) If the initial speed of the child is halved then the new height can be determined as follows:
$\frac{y_{A, new}}{y_{A,old}}=\frac{v^2_{B, new}}{v^2_{B,old}/2g}=\frac{(\frac{1}{2}v_{B,old})^2}{(v_{B,old})^2}=\frac{1}{4}$
This can be rearranged as:
$y_{A, new}=\frac{1}{4} y_{A, old}=\frac{1}{4}(0.208m)=0.0520m$
