URGENT‼
Physics Angled Projectile

5. 0.5 m/s

6. 2.06 m/s

7. 0.22 m

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

v=ir = 3x15 =45 v