At this point you should know that
if s(t) is the height obtained, then
v(t) , or velocity, is s'(t) and
a(t) , the acceleration, is v'(t) or s''(t)
Also having done some of these, by now you should know that for the above,
s(t) = -16t^2 + 64t + 480 and
v(t) = -32t + 64
so when it hits the ground, isn't s(t) = 0 ??
so
-16t^2 + 64t + 480 = 0
t^2 - 4t - 30 = 0
t = 2 ± √34 , we can reject the negative value
t = 2+√34
sub that into v(t) and you are done
use a(t) = -32 ft per second squared as the acceleration due to gravity.An object is thrown vertically downward from the top of a 480-ft building with an initial velocity of 64 feet per second. With what velocity does the object hit the ground?
This type of problem always shows up on my test and I always get it wrong. Can someone show me how to do it? Thanks.
1 answer
1 answer