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

To calculate the horizontal distance traveled by the ball before making contact with the ground, we need to analyze the horizontal and vertical components separately.

Horizontal Component:
The horizontal component of the initial velocity is given by v*cos(theta), where v is the magnitude of the initial velocity (15 m/s) and theta is the angle of projection (30 degrees).
Horizontal component (vx) = v * cos(theta) = 15 * cos(30) = 12.99 m/s.

Using the formula x = v*t, where x is the horizontal distance traveled, v is the horizontal component of the velocity, and t is the time of flight, we can solve for x by substituting the known values. Since there is no horizontal acceleration, x = (12.99 m/s) * t.

Vertical Component:
The vertical component of the initial velocity is given by v*sin(theta), where v is the magnitude of the initial velocity (15 m/s) and theta is the angle of projection (30 degrees).
Vertical component (vy) = v * sin(theta) = 15 * sin(30) = 7.5 m/s.

Using the formula h = ut + (1/2) * a * t^2, where h is the vertical displacement, u is the initial velocity (7.5 m/s), t is the time of flight, and a is the acceleration due to gravity (-9.8 m/s^2), we can solve for t.
0 = (7.5 m/s) * t + (1/2) * (-9.8 m/s^2) * t^2.
Rearranging the equation, we get -4.9t^2 + 7.5t = 0.

Solving this quadratic equation using the quadratic formula: t = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = -4.9, b = 7.5, c = 0, we find two solutions: t = 0 s and t = 1.53 s. Since t = 0 s corresponds to the initial time when the ball is thrown, we only take t = 1.53 s as the time of flight.

Substituting the value of t = 1.53 s into the equation x = (12.99 m/s) * t, we find the horizontal distance traveled by the ball before making contact with the ground:
x = (12.99 m/s) * (1.53 s) = 19.88 m.

Therefore, the ball travels approximately 19.88 meters horizontally before making contact with the ground.

To solve this problem, we can use the principles of projectile motion. Projectile motion involves the motion of an object that is thrown or launched into the air, experiencing both horizontal and vertical motion.

First, let's analyze the horizontal motion of the ball. Since there is no acceleration in the horizontal direction, the ball will travel at a constant velocity without changing its speed. Therefore, we can use the equation:

x = v * t

where x is the horizontal distance, v is the initial horizontal velocity of the ball, and t is the time it takes to reach the ground.

In this case, the initial horizontal velocity of the ball is given as 15 m/s. To find the time it takes to reach the ground, we need to analyze the vertical motion.

Now, let's examine the vertical motion of the ball. The vertical motion is influenced by gravity, which causes the ball to accelerate downward. The equation for the vertical displacement can be given as:

y = v0 * t + (1/2) * a * t^2

where y is the vertical displacement, v0 is the initial vertical velocity, t is the time, and a is the acceleration due to gravity (approximately 9.8 m/s^2).

In this case, the ball is thrown at an angle of 30 degrees to the horizontal. We can break down the initial velocity into its horizontal and vertical components using trigonometry.

The initial vertical velocity can be calculated as:

v0y = v * sin(theta)

where v is the initial velocity of the ball and theta is the angle of projection.

Given that v is 15 m/s and theta is 30 degrees, we can calculate the initial vertical velocity:

v0y = 15 * sin(30) = 7.5 m/s

Now, we can find the time it takes for the ball to reach the ground by setting the vertical displacement y equal to zero:

0 = v0y * t - (1/2) * g * t^2

Substituting the known values, we have:

0 = 7.5 * t - (1/2) * 9.8 * t^2

Simplifying the equation, we get:

4.9 * t^2 = 7.5

Dividing both sides by 4.9, we have:

t^2 = 7.5 / 4.9

Taking the square root of both sides, we get:

t = sqrt(7.5 / 4.9) ≈ 1.24 seconds

Now that we know the time it takes for the ball to reach the ground, we can substitute this value into the horizontal displacement equation:

x = v * t
x = 15 * 1.24 ≈ 18.56 meters

Therefore, the ball will travel approximately 18.56 meters horizontally before making contact with the ground.

Note: Please double-check the formulas and calculations as I provided them to ensure accuracy.