A projectile is launched with an inital velocity of 30.0m/s at an angle of 60 degrees above the horizontal. How far does it travel. Please show me how to solve this.

Draw your triangle

Decide whether to use sin or cos
find value
dh=vht
Solve for dh.

Payday: stop answering if you have no idea.

time in air: hf=hi+30sinTheta*t-1/2 g t^2
hi=hf=0
solve for time in air t.

then, hoizontal distance
distance=timeinair*30cosTheta

To solve this problem, we can use the equations of projectile motion. The horizontal and vertical components of motion are treated separately.

Let's start by finding the time it takes for the projectile to reach its maximum height. We can use the vertical motion equation:

vf = vi + gt

In this case, vf (final vertical velocity) is 0 m/s at the maximum height, vi (initial vertical velocity) is the vertical component of the initial velocity (30.0 m/s * sin(60º)), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time we are looking for.

0 = (30.0 m/s * sin(60º)) - (9.8 m/s^2 * t)

We can now solve this equation to find t.

0 = (30.0 m/s * 0.866) - (9.8 m/s^2 * t)
9.8 m/s^2 * t = (30.0 m/s * 0.866)
t = (30.0 m/s * 0.866) / 9.8 m/s^2
t ≈ 2.58 s

So it takes approximately 2.58 seconds for the projectile to reach its maximum height.

Next, let's find the total horizontal distance traveled by the projectile. We can use the horizontal motion equation:

d = vi * t

In this case, vi (initial horizontal velocity) is the horizontal component of the initial velocity (30.0 m/s * cos(60º)), and t is the time we just calculated.

d = (30.0 m/s * cos(60º)) * 2.58 s

Now we can solve for d.

d = (30.0 m/s * 0.5) * 2.58 s
d ≈ 38.7 m

Therefore, the projectile travels approximately 38.7 meters.