Question

The former Sears Tower in Chicago is 443m tall. Suppose a book is dropped from the top of the building. What would be the book's velocity a point 221m above the ground?

Answer 1

here we can use energy conservation

like initial kinetic + potential energy is always conserved and it will be same at all points

so we can say

$KE_i + PE_i = KE_f + PE_f$

$\frac{1}{2}mv_i^2 + mgh_1 = \frac{1}{2}mv_f^2 + mgh_2$

now we can plug in all the given values in it

$v_i = 0$

$h_1 = 443 m$

$h_2 = 221 m$

$\frac{1}{2}m*0 + m*9.8*443 = \frac{1}{2} m*v_f^2 + m*9.8*221$

now divide whole equation by mass "m"

$9.8*443 = \frac{1}{2} v_f^2 + 9.8*221$

$2175.6 = \frac{1}{2}v_f^2$

$v_f = 65.96 m/s$

so final speed will be 65.96 m/s