Question

stone that is dropped from the top of building strikes the ground in 10 seconds. A. Compute the height of the building in feet. B. What is the velocity of the stone at the end of the 7th second?

Answer 1

Answer: 1607.6 ft, 68.6 m/s

when a stone is dropped from height (h), its initial velocity is u=0. It strikes the ground in time, t=10 s.

From the equation motion, we can compute the height of the building:

$h=ut+\frac{1}{2}at^2$

a= g=9.8 m/s²

substitute the values:

h=0+0.5×9.8 m/s²×(10s)²=490 m

now, 1 m= 3.23 ft

⇒490 m=1607.6 ft

Hence, the height of the building in feet is 1607.6 ft

The velocity at the end of the 7th second can be calculated by the following equation of motion:

v=u+at

⇒v=0+9.8 m/s²×7 s = 68.6 m/s.

Hence, the velocity of the stone at the end of the 7th second is 68.6 m/s.