A baseball is thrown vertically upward with an initial velocity of 14 mps. Find the: a) velocity with which it strikes the ground b) maximum height it can reach c) time taken to reach that height

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The movement of a body thrown vertically upwards is uniformly accelerated (or decelerated, if you wish). The speed `V` of a body is equal to `V(t) = V_0 - gt,` where `t` is a time, `V_0` is an initial upward speed and `g approx 9.8 m/s^2` is the gravity acceleration. The minus sign before `g` reflects the fact that gravity acceleration is directed downwards.

The height of a body is `H(t) = H_0 + V_0 t - (g t^2)/2,` where `H_0` is the initial height (zero in our problem). A baseball strikes the ground at `t_1gt0` such that `H(t_1) = 0.` It is obvious that `t_1 = (2 V_0)/g.` The velocity at this moment is `V(t_1) = V_0 - g*(2 V_0)/(g) = -V_0,` i.e. the same but downwards.

The maximum height is reached at the parabola vertex, and the time is `t_2 = -b/(2a) = V_0/(g) approx 1.43 (s)` (the half of `t_1,` actually). The height at that time is `(V_0)^2/(2g) approx 10 (m).`

The answers: a) `14 m/s` downwards, b) `10 m,` c) `1.43 s.`


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