Answer:
the angle is about 67.79 degrees
Explanation:
We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)
We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

we can use this information to find the y component of the velocity at that height via the formula:

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

The number of ocean waves that pass a buoy in one second is the frequency of the <span>wave. The crest of a transverse wave is its highest point. </span>
It would last as long as the applied force continued, or until the accelerating object hit something.
Answer:
Sound waves are reflected back
Answer:
1) No, the car does not travel at constant speed.
2) V = 9 ft/s
3) No, the car does not travel at constant speed.
4) V = 5.9 ft/s
Explanation:
In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:
V(t) = 2*t + 2 Since the speed depends on time, the speed is not constant at any time.
For the average speed we evaluate the formula for t=2 and t=5:
d(2) = 8 ft and d(5) = 35 ft

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:
d(1.8) = 6.84 ft and d(2.1) = 8.61 ft
