Answer:

For the first problem:

Given:

- Initial velocity (vi) = 6.2 m/s

- Angle (θ) = 52 degrees

a) To solve for the time it took the ball to land, we can use the formula:

t = (2vi sin θ)/g, where g is the acceleration due to gravity (9.81 m/s^2)

Substituting the given values, we get:

t = (2 x 6.2 x sin 52)/9.81

t ≈ 1.2 seconds

Therefore, the ball took about 1.2 seconds to land.

b) To solve for the height the ball flew, we can use the formula:

h = (vi^2 sin^2 θ)/(2g)

Substituting the given values, we get:

h = (6.2^2 x sin^2 52)/(2 x 9.81)

h ≈ 1.7 meters

Therefore, the ball flew about 1.7 meters high.

c) To solve for the range of the ball, we can use the formula:

R = (vi^2 sin 2θ)/g

Substituting the given values, we get:

R = (6.2^2 sin 104)/9.81

R ≈ 9.2 meters

Therefore, the range of the ball is about 9.2 meters.

For the second problem:

Given:

- Initial velocity (vi) = 24 m/s

- Angle (θ) = 28 degrees

a) To solve for the time the ball is in the air, we can use the formula:

t = (2vi sin θ)/g, where g is the acceleration due to gravity (9.81 m/s^2)

Substituting the given values, we get:

t = (2 x 24 x sin 28)/9.81

t ≈ 2.6 seconds

Therefore, the ball is in the air for about 2.6 seconds.

b) To solve for the distance the ball travels before it hits the ground, we can use the formula:

R = (vi^2 sin 2θ)/g

Substituting the given values, we get:

R = (24^2 sin 56)/9.81

R ≈ 131 meters

Therefore, the ball travels about 131 meters before it hits the ground.

c) To solve for its maximum height, we can use the formula:

h = (vi^2 sin^2 θ)/(2g)

Substituting the given values, we get:

h = (24^2 x sin^2 28)/(2 x 9.81)

h ≈ 32 meters

Therefore, the maximum height of the ball is about 32 meters.

Explanation:

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