Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8
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<u>Answer:</u>
15.97 N force is tending to pull Rover forward
<u>Explanation:</u>
The woman pulls on the leash with a force of 20.0 N at an angle of 37° above the horizontal. The arrangement is shown in the given figure,
We nee to find the pulling force P. The 20.0 N force has two components, 20.0 cos 37 in horizontal direction and 20.0 sin 37 in vertical direction.
The horizontal component is equal to pulling force P, which will pull Rover forward/
So, P = 20.0 cos 37 = 15.97 N
15.97 N force is tending to pull Rover forward.