Answer
$1.1m$
Work Step by Step
We can find the required distance $d$ as follows:
According to the law of conservation of energy
$mgh_i+\frac{1}{2}mv_i^2=mgh_f+\frac{1}{2}mv_f^2$
We plug in the known values to obtain:
$(1.9Kg)(9.8m/s^2)(1.5m)+\frac{1}{2}(1.9Kg)(0m/s^2)=(1.9Kg)(9.8m/s^2)(0.25m)+\frac{1}{2}(1.9Kg)v_f^2$
This simplifies to:
$v_f=4.95m/s$
The vertical distance is given as
$y=y_{\circ}+v_{\circ y}t+\frac{1}{2}a_yt^2$
We plug in the known values to obtain:
$0=y_{\circ}+0(t)+\frac{1}{2}a_yt^2$
This simplifies to:
$t=\sqrt{\frac{-2y_{\circ}}{a_y}}$
We plug in the known values to obtain:
$t=\sqrt{\frac{-2(0.25m)}{-9.81m/s^2}}$
$t=0.226s$
Now $d=x_{\circ}+v_{\circ x}t+\frac{1}{2}a_xt^2$
$d=(0m)+(4.95m/s)(0.226s)+\frac{1}{2}(0m/s^2)(0.226s)^2$
$d=1.1m$
