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.