More Homework Fun: Physics Projectile Motion Problem

[JohnCU]

A ball is projected from ground level. Later, 4.5 seconds after being projected it strikes the top of a wall 6.8 m above ground level at a horizontal distance of 64.0 m from the point of projection. What was the initial speed of the ball in m/s?

Okay, so the max height is 6.8m and the range is 64.0m...but I don't have an angle and all my equations need angles...

Btw, the answer is 28m/s, how you get 28m/s is the question.

 

[minendo]

n/m

 

[ClueLis]

Call "a" the angle between the ground and the initial trajectory of the ball and v the initial speed. We know that, horizally, the speed is 64m/4.5s=14.22m/s. So v*sin(a)=14.22m/s.

Now, it takes 4.5s for the ball to go up, and come down to 6.8m. So you set up the displacement function:

y(t)=-1/2(9.8)t^2+v*cos(a)*t
y(4.5)=6.8m

Once you have v*cos(a)*t and v*sin(a), finding v should be pretty easy.

 

[theApp]

Ok, here goes. I'm a Physics noob, so I dunno if this is right....

You want to find out the initial velocity in the x and y directions...
Lets start with the y direction.

Ok, we don't have an initial velocity or a final, but we do have time (4.5s), acceleration (-9.80m/s**2), and a displacement (6.8m)

Now, we need an equation without a velocity final in it, so we are going to use:
y=vi*t+1/2at^2

Plugging in and solving we get the initial velocity in the y direction to be 23.6 m/s

Ok, so the we have the initial y. Now lets find the initial x velocity. This one is easy as the x is at a constant velocity. So this is just going to be the displacement in the x direction over the time; 64 m / 4.5 s = 14.2 m/s

Now, we have both the initial x and y velocitys and using the Pythogorean (sp) thereom, we can find the initial overall velocity.

SQRT(23.6^2+14.2^2) = 27.5 m/s or 28 m/s rounded.

I am kinda sure that is how this works, but I've only been doing it for about a day, so someone who is better versed in Physics will probably correct me. Good luck.

 

[JohnCU]

Thanks!