A ball is thrown in the air at an angle of 40∘. If the maximum height it reaches is 10m, what must be its initial speed?

Let v = the launch velocity.

Because the launch angle is 40° (with the horizontal), the initial vertical velocity is

v * sin (40°) = 0.6428v m/s.

Assume g = 9.8 m/s² and ignore air resistance.

At maximum height, the vertical velocity is zero.

Because the maximum vertical height is 10 m, therefore

(0.6428v m/s) ² - 2 * (9.8 m/s²) (10 m) = 0

0.4132v² = 196

v = 21.78 m/s

Answer: 21.8 m/s (nearest tenth)