A stone is dropped from the top of a building such that its height, in feet, is given by s ( t ) = − 16 t^2 + s_0. It makes impact with the ground with a velocity of − 145 ft/s .

Question

A stone is dropped from the top of a building such that its height, in feet, is given by s ( t ) = − 16 t^2 + s_0.

It makes impact with the ground with a velocity of − 145 ft/s .

Answer 1

s(t) = -16t^2 + so

so = the initial height

v(t) =-145 when s(t)=0

v(t) = -32t = -145

t = 145/32 = about 4 1/2 seconds when it hits the ground when velocity = -145

s(145/32) = - 16(145/32)^2 + so =0

so =16(145/32)^2 = 145^2/64

so = about 328 1/2 feet = initial height= height of the building

general equation is

s(t) = (a/2)t^2 + vot +so

where vo= initial velocity= 0 in this example

so=intitial height= about 328.5 feet in this example

a = accelertion = the effect of gravity =-32 feet per second per second

Answer 2

There are two things we could ask: how long did the stone fall, and what is the height of the building.

We know the final velocity (-145 ft/s) so it's helpful to differentiate the position function s(t) to obtain the velocity function v(t) = s'(t) = -32t. If t_1 is the time when the stone hits the ground, then v(t_1) = -145 = -32t_1, by our assumption that the stone makes impact with a velocity of -145 ft/s. Dividing we obtain t_1 = 145/32 = 4.53125 seconds.

Thus the stone fell for about 4.53125 seconds. (Note: this could be done without calculus, because in free-fall an object accelerates downward at 32 feet per second per second, which gives us that v(t) = -32t.)

The next question: how tall is the building. We know that the position (height) of the stone at t_1 is 0, because that is when it makes impact. Hence s(t_1) = 0 = -16t^2 + s_0, where s_0 = s(0) is the initial height of the stone. Solving this equation: s_0 = 16(4.53125)^2 = 328.515625. The building is about 328.52 feet tall.