Answer:
5. 0.5 m/s
6. 2.06 m/s
7. 0.22 m
Step-by-step explanation:
From the question given above, the following data were obtained:
Horizontal range (Rₓ) = 0.21 m
Time of flight (T) = 0.421 s
Time (t) to reach the top = 0.2105 s
5. Determination of the horizontal velocity of the ball.
Horizontal range (Rₓ) = 0.21 m
Time of flight (T) = 0.421 s
Horizontal velocity (vₓ) =?
Rₓ = vₓ × T
0.21 = vₓ × 0.421
Divide both side by 0.421
vₓ = 0.21 / 0.421
vₓ = 0.5 m/s
Thus, the horizontal velocity is 0.5 m/s.
6. Determination of the initial vertical velocity of the ball.
Time (t) to reach the top = 0.2105 s
Final vertical velocity (vբᵧ) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Initial vertical velocity (vᵢᵧ) =?
vբᵧ = vᵢᵧ – gt (ball is going against gravity
0 = vᵢᵧ – (9.8 × 0.2105)
0 = vᵢᵧ – 2.06
Collect like terms
0 + 2.06 = vᵢᵧ
vᵢᵧ = 2.06 m/s
Thus, the initial vertical velocity of the ball is 2.06 m/s.
7. Determination of the maximum height reached by the ball.
Time (t) to reach the top = 0.2105 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =?
h = ½gt²
h = ½ × 9.8 × 0.2105²
h = 4.9 × 0.2105²
h = 0.22 m
Thus, the maximum height reached by the ball is 0.22 m