Agolfer hits a shot to a green that is elevated 2.90 m above the point where the ball is struck. the ball leaves the club at a speed of 19.2 m/s at an angle of 50.0˚ above the horizontal. it rises to its maximum height and then falls down to the green. ignoring air resistance, find the speed of the ball just before it lands. v = entry field with incorrect answer now contains modified data entry field with correct answer

Refer to the diagram shown below.

The horizontal component of the launch velocity is
vx = (19.2 m/s) cos 50° = 10.4132 m/s
The vertical component of the launch velocity is
vy = (19.2 m/s) sin 50° = 12.4099 m/s

Assume that air resistance is negligible.
Take g = 9.8 m/s², the acceleration due to gravity.

Let Vy = vertical component of the arrival velocity, and
      Vx = horizontal component of arrival velocity.

Because air resistance is ignored,
Vx = vx = 10.4132 m/s

The vertical component of the arrival velocity at a height of h = 2.9 m is given by
Vy² = (12.4099 m/s)² - 2*(9.8 m/s²)*(2.9 m) = 97.1656 (m/s)²
Vy = 9.8573 m/s

The arrival velocity is the vector sum of Vx and Vy. Its magnitude is
V = √(10.4132² + 9.8573²) = 13.6706 m/s

The angle that V makes below the x-axis is
tan⁻¹ (9.8573/10.4132) = 43.4°

The speed of the ball just before it lands is 13.7 m/s, at an angle of 43° measured clockwise from the horizontal x-axis.

Answer: buoyancy forceexplanation: in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.