The body is thrown with speed. 10 m / s at an angle of 45 ° to the horizontal.

The body is thrown with speed. 10 m / s at an angle of 45 ° to the horizontal. Determine the height of the body at the moment when its x coordinate becomes 3 m.

The origin of the coordinate system is at the point from which the throw was made. Y coordinate (height h) – directed up.
Let’s split the speed into vertical and horizontal components:
Vv = V * sin45 °;
Vg = V * cos45 °.
The horizontal speed will be constant. The dependence of the X coordinate on time:
X = Vg * t = V * cos45 ° * t.
Substitute the values from the condition and find the time of flight 3 m horizontally:
t = X / V * cos45 ° = 3 m / (10 m / s * (√2 / 2)) = 0.424 s.
At the moment t = 0.424 s, the body will be at the height h:
h = V * sin45 ° * t – gt² / 2 = 10 m / s * (√2 / 2) * 0.424 s – 0.5 * 10 m / s² * 0.424² s² = 2.1 m.
Answer: 2.1 m.