Posted by **sean** on Saturday, October 7, 2006 at 5:52pm.

a ball is thrown at 15m/s at an angle of 30 degrees to the horizontal , how far does it travel horizontally before making contact with the ground?

I don't remember very well, but I believe the formula you need is:

x0 = -.5at^2 + vt + x

where x0 is the initial distance (a.k.a., 0)

a = acceleration (m/s^2)

v = velocity

t = time (seconds)

x = distance (m)

For this problem, you need to split it into the horizontal and vertical.

Horizontally you have no acceleration so your formula becomes (if I remembered it correctly):

x0 = vt + x

or

0 = (15t) + x

You want to find x so now you know you have to find t.

Vertically, you have:

0 = -.5(9.8)t^2 + (0)t + h

where h = height

h can be figured out by doing:

sin(30) = h/15

so we find h = 7.5m

so now we know that:

4.9t^2 = 7.5

t^2 = 1.530612245

t = 1.237179148

t ~ 1.24s

So, earlier we had:

0 = (15t) + x

so now that we know t

x = 18.55768722m

or in other words

x ~ 18.56m

I hope that was helpful. If I remembered the formula incorrectly, just change the formula accordingly. :)

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