From a spring pistol at a distance of 2 m above the ground, a shot was fired vertically at the top.

From a spring pistol at a distance of 2 m above the ground, a shot was fired vertically at the top. The ball moves at a speed of 5 m / s. Determine the lift height and how long the ball moves

Given:

h1 = 2 meters – starting height;

V = 5 m / s – initial ball velocity;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to determine the height of the ball rise H and the time of its motion t.

Since the ball is moving upward, and at the maximum point of ascent its speed will be equal to 0, we find the time of the ball’s flight to the maximum point:

t1 = V / g = 5/10 = 0.5 seconds.

During this time, the ball will cover the distance:

h2 = V * t – g * t ^ 2/2 = 5 * 0.5 – 10 * 0.5 * 0.5 / 2 = 2.5 – 1.25 = 1.25 m.

Since the ball has already started flying from the height h1, the maximum height H will be equal to:

H = h1 + h2 = 2 + 1.25 = 3.25 meters.

Now let us find the time of the fall of the ball from the height H:

t2 = square root (2 * H / g) = square root (2 * 1.25 / 10) =

= square root (0.25) = 0.5 s.

The total time of movement will be equal to the time of the upward movement + the time of the fall:

t = t1 + t2 = 0.5 + 0.5 = 1 s.

Answer: the height of the ball rise is 3.25 meters, the total movement time is 1 second.