The body was thrown at an angle α = 15 degrees to the horizon with an initial velocity v0 = 20 m / s.

The body was thrown at an angle α = 15 degrees to the horizon with an initial velocity v0 = 20 m / s. At what distance L from the throwing point will it fall on the horizontal surface of the Earth? Disregard the influence of air.

Given:

v0 = 20 meters per second – initial speed of a body thrown at an angle to the horizon;

a = 15 ° – the angle at which the body was thrown;

g = 10 m / s2 – acceleration of gravity.

It is required to determine L (meter) – at what distance from the throwing point the body will fall.

Let us find the vertical component of the speed, which is responsible for the flight duration:

vv = v0 * sin (a) = 20 * sin (15) = 20 * 0.26 = 5.2 meters / second.

Then the total flight time will be:

t = 2 * (vw / g) = 2 * (5.2 / 10) = 10.4 / 10 = 1.04 seconds.

The flight range will be equal to:

L = vg * t = v0 * cos (a) * t = 20 * 0.97 * 1.04 = 19.4 * 1.04 = 20.2 meters (the result has been rounded to one decimal place).

Answer: the flight range of the body will be equal to 20.2 meters.