i do believe the answer is c not sure but correct me if im wrong
Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance. Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement. Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m
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Answer:
The new speed of the ball is 176.43 m/s
Explanation:
Given;
mass of the ball, m = 7 kg
initial speed of the ball, u = 5 m/s
applied force, F = 300 N
time of force action on the ball, t = 4 s
Apply Newton's second law of motion;

where;
v is new speed of the ball

Therefore, the new speed of the ball is 176.43 m/s