The ball was thrown at an angle of 30 ‘to the horizon at a speed of 20 m / s.

The ball was thrown at an angle of 30 ‘to the horizon at a speed of 20 m / s. To what maximum height it will rise. How far from the point of throw will the ball fall?

V = 20 m / s.

∠α = 30 °.

g = 9.8 m / s ^ 2.

h -?

S -?

The ball will move horizontally evenly in a straight line at a speed V * cosα. The ball will move vertically with gravitational acceleration g and initial velocity V * sinα.

The flight range S is found by the formula: S = V * cosα * t, where t is the flight time of the ball.

h = (V * sinα) ^ 2/2 * g.

h = (20 m / s * sin30 °) ^ 2/2 * 9.8 m / s ^ 2 = 5.1 m.

Let’s find the time of lifting the ball t1 to the maximum height h by the formula: t1 = V * sinα / g.

t1 = 20 m / s * sin30 °) ^ 2 / 9.8 m / s ^ 2 = 1.02 s.

The entire flight time t will be: t = 2 * t1.

t = 2 * 1.02 s = 2.04 s.

S = 20 m / s * cos30 ° * 2.04 s = 35 m.

Answer: h = 5.1 m, S = 35 m.