Question

85. Standing at the edge of a building of height 9.2 m, a woman reaches out over the edge and throws a penny vertically upward. It rises, then falls past her, hitting the ground below 2.5 s after release. What height did the penny attain? What was its initial velocity?

Answer 1

Time taken for the penny to reach ground is $t=2.5\,s$

Distance traveled by penny in the vertical direction is $y=ut+\frac{1}{2}at^2$

In the downward direction, $y=-9.2\,m$ , $a=-9.81\,m/s^2$

$-9.2=u(2.5)-\frac{1}{2}*9.81*(2.5)^2$

$u=8.58\,m/s$

Initial velocity of penny is $u=8.58\,m/s$

Maximum height reached by penny is $H=\frac{u^2}{2g}=\frac{8.58^2}{2*9.81}=3.75\,m$ from the point of projection.

From the ground, height reached by penny is $H'=9.2+3.75=12.95\,m$