Answer:
5.09 m/s
Explanation:
Use the height to find the time it takes to land:
y = y₀ + v₀ᵧ t + ½ gt²
0 = 8.0 m + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.28 s
Now use the horizontal distance to find the initial velocity.
x = x₀ + v₀ₓ t + ½ at²
6.5 m = 0 m + v₀ (1.28 s) + ½ (0 m/s²) (1.28 s)²
v₀ = 5.09 m/s
Answer:
3) D: 31 m/s
4) D: 84.84 metres
Explanation:
3) Initial velocity along the x-axis is;
v_x = v_o•cos θ
Initial velocity along the y-axis is;
v_y = v_o•sin θ
Plugging in the relevant values, we have;
v_x = 31 cos 60
v_x = 31 × 0.5
v_x = 15.5 m/s
Similarly,
v_y = 31 sin 60
v_y = 31 × 0.8660
v_y = 26.85 m/s
Thus, magnitude of the initial velocity is;
v = √(15.5² + 26.85²)
v ≈ 31 m/s
4) Formula for horizontal range is;
R = (v² sin 2θ)/g
R = (31² × sin (2 × 60))/9.81
R = 84.84 m
Answer:
Explanation:
The force of friction between the quails feet and the ground is:
So the coefficient of static is solve
The answer to this question is 3.69
