Given an initial height, what is the equation to find velocity at impact?

Question

Assuming g=32.174 ft/sec/sec in a vacuum and no time given. i.e., what is Vf after distance, given V0 and a?

I think this involves a calculus integral, so, assuming an object dropped from 4000 ft. in a vacuum, what is t and Vf?

I think this involves a calculus integral, so assuming an object dropped from 4000 ft. in a vacuum, what is t and Vf? Also, what is t when V=60mph?
(g=32.174 ft/sec/sec)
(you might recognize this from a Lexus commercial)

Answer 1

Velocity is the first derivative of distance travelled with respect to time and acceleration is it's second derivative. Working backward and integrating as we go, we can obtain expressions for velocity and distance relative to time.
V=gt+U.........(1) { g integrated = gt+constant}
S=1/2gt^2+Ut +K....(2) {gt+U integrated = 1/2gt^2+Ut + another constant} at time zero distance is zero, so K=0;
Likewise, at time zero, V=0,so U =0;
So, we can simplify are two equations as :
V=gt ; S= 1/2gt^2
How long will the object take to reach 4000ft? sqr(8000/32.174)sec=15.7686secs
What velocity will it have? 32.174*15.7686ft/sec =507.34ft/sec.

Answer 2

using the equation
v^2 =u^2 +2as u=0 since body is dropped
v^2=2as
v=sqrt2as
v=sqrt(2*32.174*4000)
v=507.3381515320920550 ft/sec
v=507.338 ft/sec
(to 3 d.p)
thus the velocity the body has just before impact is 507.338ft/sec taking vector directions downwards as positive.
using s=ut+1/2at^2, u=0
t=sqrt(2s/a)
t=sqrt(2*4000/32.174)
t=15.76sec
thus time elapsed befor hitting ground=15.76 sec
1 mile=5280 ft
60 mph=(60*5280)/(60*60)
=88ft/sec
using v=u+at,
t=v/a
=88/32.174
=2.73 secs.
so, when the velocity is 60 mph,the time elapsed would be 2.73 secs.

Answer 3

You can find time from the acceleration, initial velocity, and distance fallen, but it is not necessary. Solve this equation for vf and you should get your answer. Make sure that your units are all consistent as well.

vf^2 - v0^2 = 2*a*x

Answer 4

The ecuation for the object's motion is h = g * t * t / 2 . At impact h = a, and you get the time: t = sqrt (2a/g). The velocity is v = g * t = sqrt (2*g*a).