Help...pHysics
posted by sean on .
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. :)

7.5 is the initial velocity in the x direction so it isn't the height the max height it reaches is 8.61m d=19.88