Physics, 31.01.2021 02:00 isabellatessa86

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Physics Angled Projectile

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Answer from: emilylittle2442

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

Answer from: Quest

the following is queen elizabeth’s purposes:

- to encourage the parliament that the technique of picking successors to the seat based merely on birthright needs to be reviewed.

- (not) to convince the parliament that her kid should not be measured as a likely choice for heir to the english power.

- to persuade the parliament that merely since she has a child does not assure that the kid will be a capable leader.

- to sway the parliament how shattering it could be if they proceeded discourteously concerning her or her child.

Answer from: Quest

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

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Physics Angled Projectile
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