# Homework Help: Final velocity in SUVAT equations -- ball thrown from a wall

1. Apr 15, 2015

### gracy

1. The problem statement, all variables and given/known data:There is a wall of height 30 meters.From the top edge of the wall a ball is thrown up with a velocity of 5m/s.Find the time taken by the ball to come to the base of the wall.
Take g=10 m/s^2

2. Relevant equations:I used
V=u+at

3. The attempt at a solution
v=0 as ball will come to rest when it will reach at base of the wall.
u=5 m/s
a= - 10 m/s^2( negative because of chosen coordinate system)
so time come out to be 0.5 s
But it is wrong it should be 3s.
I think I am going wrong in taking final velocity equal to zero.
So I should rather use s=ut +1/2at^2

2. Apr 15, 2015

### haruspex

Yes, the final velocity is not zero. The SUVAT equations are only valid for constant acceleration. When the ball hits the ground it undergoes a completely different acceleration, the the equations are only valid up to an instant before that.
In this case, you do not know the final velocity, and do not need to determine it. There are five variables in the SUVAT equations, and each of the five equations involves four of them. Typically, you know the values of three and wish to find a fourth. List what those four variables are and select the equation that includes them.

3. Apr 15, 2015

### gracy

S= - 30 meters
u=5 meters
a=- 10 m/s^2
-30=5t+1/2 (-10)t^2
-30 =5t-5t^2
0=-5t^2+5t+30
it takes a shape of quadratic equation.
so will have to use below formula

t= - 5 ± √(5)^2-4 (-5)30/2(-5)
t=-5 ± √625/(-10)
t=-5 ±25/(-10)
t=3 or -2
time can not be negative hence t=3s

Right?

4. Apr 15, 2015

### haruspex

That's all correct, but you should be more careful with parentheses. E.g. your expression - 5 ± √(5)^2-4 (-5)30/2(-5), taken literally, says - 5 ± (√(5))^2-((4 (-5)30/2)(-5)), giving a very different result. You should have written (- 5 ± √(5^2-4 (-5)30))/(2(-5)).
Also, you could have made it a bit easier by first simplifying 0=-5t^2+5t+30 to t^2-t-6=0. Then you might have spotted the factorisation (t-3)(t+2).
Note the physical significance of the -2 solution. It means that if a ball were thrown up from the ground such that after 2 seconds it reaches the top of the wall then it will at that point have an upward speed of 5m/s, and from that point on it will look exactly like the given problem.

5. Apr 18, 2015

### gracy

If an object is thrown from certain height with horizontal velocity,it's final (when it touches the ground)will be zero,right?
But when we use suvat equation V sub y =u sub y +a sub y t
u sub y=0
V sub y =a sub y t
as a sub y=9.8 m/s^2 (taking downward direction positive)
t= time of flight which is also non zero.
V sub y can not be zero.Right?

6. Apr 18, 2015

### Staff: Mentor

That's right, it can't be zero. Under fixed acceleration of 10m/s2 once released with vy = 0m/s its velocity can only increase, it cannot again become 0m/s.

As haruspex explained, to be brought to standstill it must be acted on by a different acceleration, and you have not performed calculations with acceleration other than 10m/s downwards.

7. Apr 18, 2015

### gracy

But I am not saying that object will be standing still,there may be some final horizontal velocity,I am concern about final vertical velocity.Is it also non zero?

8. Apr 18, 2015

### Staff: Mentor

The horizontal and vertical motions can be considered independently, so a "vertical standstill" need not imply a "horizontal standstill". I was considering only vertical motion.

The horizontal component of velocity is unaffected by gravity, though we all know that once the projectile encounters the ground conditions instantly change with the result that both of its velocity components will almost immediately drop to 0m/s.

9. Apr 18, 2015

### gracy

But according to
V=u+at
The velocities can not be zero.

10. Apr 18, 2015

### Staff: Mentor

That's right, the equations say that speed must go on increasing forever and without limit, and it will UNLESS the acceleration changes from 10m/s2.

It is obvious that contact with the ground causes a new force to suddenly enter into the picture, and this new force introduces a new acceleration in opposition and much greater in magnitude than gravity. So the former equations would need to be updated with the new data to account for the observed behaviour if it were important that we continue to model the projectile's path after contact with the ground.

11. Apr 18, 2015