If an object is thrown downward with an initial velocity of 30 feet per second from a height of 80 feet above the ground how long will it take to hit the ground?

Answer 1

The solution is as follows. First find the final velocity. Second use the final velocity to find the time.

(final velocity squared) = (initial velocity squared) + 2gh

(vf)^2=(vi)^2+2gh

(vf)^2=(30ft/s)^2+2(32ft/s^2)(80ft)

(vf)^2=900(ft/s)^2+5120(ft/s)^2=6028(ft/s)^2

taking the square root yields

vf=77.59(ft/s)

The final velocity and the time are related by the equation vf=vi+gt, rearranging

t=(vf-vi)/g

t=[77.59(ft/s)-30(ft/s)]/32(ft/s^2=47.59(ft/s)/32ft/s^2=1.49s You can check the answer as follows:

h=vi(t)+(1/2)gt^2

h=30(ft/s)(1.49s)+(1/2)(32(ft/s^2)(1.49s)^2

h=44.7ft+16(ft/s^2)[2.22s^2)

h=44.7ft+35.7ft=80.2ft

which is equivalent to the 80 ft in the original problem except for rounding errors. Note: the notation here is awkward but I haven't found an equation editor iw Wiki.

h^2 means h squared

ft/s^2 means feet per second squared