The ball launched vertically upwards hit the ground after 4 seconds. Find the initial velocity of the ball and the distance traveled in 4 seconds. Acceleration of gravity is taken equal to 10 m / s squared, the influence of air is considered negligible.
Let the origin coincide with the point where the ball was launched.
The dependence of the height h (upward direction) on time is described by the equation:
h = V₀ * t – gt² / 2,
where V₀ is the initial velocity, t is the time, g is the acceleration of gravity.
After 4 seconds, the ball returned to the launch point, (h = 0):
0 = V₀ * 4 s – (10 m / s² * 16 s²) / 2;
V₀ * 4 c = 80 m;
V₀ = 20 m / s.
Let us find the maximum height h₁ at which V = 0:
V = V₀ – gt;
0 = 20 m / s – 10 * t₁;
t₁ = 2 s.
h₁ = 20 m / s * 2 s – (10 m / s² * 4 s²) / 2 = 20 m.
The full path is equal to twice the height h₁:
S = 2h₁ = 40 m.
Answer: Initial speed – 20 m / s, full path – 40 m.
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