Answer:
X = 15.88 m
Explanation:
Given:
Initial Velocity V₀ = 13.4 m/s
θ = 30.1 °
g = 9.8 m/s²
To Find horizontal distance let "X" we have to time t first.
so from motion 2nd equation at Height h = 0
h = V₀y t + 1/2 (-g) t ² (ay = -g)
0 = 13.4 sin 30.1° t - 0.5 x 9.81 x t² (V₀y = V₀ Sin θ)
⇒ t = 1.37 s
Now For Horizontal distance X, ax =0m/s²
X = V₀x t + 1/2 (ax) t ²
X = 13.4 m × cos 30.1° x 1.37 s + 0
X = 15.88 m
Well, you didn't ask a question, and 4 m/s is not an angular speed.
So all I can offer is a couple of observations:
1). The tension in the rope is
M V² / R = (1.5 kg) x (4 m/s)² / R
= (24 kg-m²/s²) / (distance of the ball from the pole).
2). Tetherball was the only thing I played at camp,
more than 60 years ago, and I loved it !
It was a tough game, because we had to skin
our own T.Rex and use his hide to make the ball
and his guts for the rope.
Answer:
The angular speed of rolling ball is 3.23 rad/sec.
Explanation:
Given that,
Radius = 1.95 cm
Speed = 6.30 cm/s
Suppose, we need to find the angular speed of rolling ball
We need to calculate the angular velocity
Using formula of angular velocity

Where, r = radius
= angular velocity
Put the value into the formula



Hence, The angular speed of rolling ball is 3.23 rad/sec.
If you mean what types of measurements are used to measure in outer space, I believe it’s light-year, astronomical unit and intergalactic measurements, but I’m not completely sure.
Hopefully this helps...