What is the velocity at impact?

[scotthands]

Homework Statement

A body is fired vertically upwards with an initial velocity of 100m/s.
(a) how far will it rise
(b) how long will it take to reach maximum height?
(c) what is the time taken before it strikes the ground again.
(d) what is the velocity at impact

Homework Equations

s= ut
v= u+at
s=ut + 0.5*at^2
v^2=u^2+2as

The Attempt at a Solution

I can't even start it, I've got no idea, it doesn't seem that the question is even giving me enough information.

 

[OJones]

In fact, assuming the system is on Earth, you need very few given variables to answer these mechanics problems. In your case, you have u, a and v. This allows you to apply the fourth equation in your list to find the height reached. From there, work through the other parts of the problem systematically until you know all the values required.

 

[scotthands]

right, so i use: v^2=u^2+2as

u=100m/s
a=?
v?

I don't no what to use for a or v

 

[OJones]

The crucial point is that the system is on Earth, i.e. the acceleration of the body is the acceleration due to gravity, which has a given value anywhere on our planet (at least to 4 or 5 sig. fig.). Use this value, which your mechanics teacher should have covered.

As for v, well, what is the question asking? It asks the maximum height reached. Maximum height implies that the object is no longer moving upward, nor has it started to come back down. At the instant when the object is at maximum height therefore, the object (not moving up or down) has speed...?

 

[scotthands]

Wait i think this is it:
u=100m/s
a=9.81
v=0

I'm not sure about the acceleration.

 

[scotthands]

ok:
A body is fired vertically upwards with an initial velocity of 100m/s.
(a) how far will it rise
(b) how long will it take to reach maximum height?
(c) what is the time taken before it strikes the ground again.
(d) what is the velocity at impact
---------------------------------------------
(a) 509.7m
(b) 10.2s

i'm not sure what equation to use for (c):
for part (c):
u=0
a=9.81
s=509.7
v=? 0 mybe?

i've tried this:
s= ut + 0.5*a*t^2
s= 0.5*a*t^2
t= sqaure root of (509.7/4.905)

and i get the answer 10.2

bit in the answer section of my book it sais 20.4 which is double, why?

 

[scotthands]

anyone?

 

[OJones]

OK, so you have the time taken to reach maximum height. In that case, the TOTAL time of the trajectory (c) is the time taken to go up AND come back down. It should be apparent that no actual calculation is really required here, but if the solution is not obvious, just do a second calculation as you did for (b) but with a straight swap for u and v where the body starts at the top at rest, and hits the ground with the same speed it started with (which unless energy is lost to the air due to frictional forces, it must do by conservation of energy). Then add time up to time down.

(P.S. you do indeed have the correct value for a, and it is VERY important that you know this value of the acceleration due to gravity, g, off by heart for many of the classical mechanics problems you will encounter).