Question

May

Answer 1

Answer:

a)  t = 0.90 s, b)  t = 0.815 s, c)  t = 0.90 s, d)  x = 3.6 m, e)  t = 0.639 s

Explanation:

all these exercises are about kinematics

a) The body is released from rest,  

           y = y₀ + v₀ t - ½ g t²

in this case when reaching the ground y = 0 and its initial velocity is vo = 0

           0 = y₀ + 0 - ½ g t²

           t² = 2 y₀ / g

           t² = 2 4 /9.81

           t² = 0.815

            t = √0.815

           t = 0.90 s

b) It is thrown upwards at v₀ = 4 m / s

         y = y₀ + v₀ t - ½ g t²

in this case the initial and final height is the same

        y = y₀ = 0

        0 = v₀ t -1/2 g t²

        t = 2 v₀ / g

        t = 2 4 /9.81

        t = 0.815 s

c) the ball is at y₀ = 4 m and its initial velocity is horizontal v₀ = 4 m / s

        y = y₀ + v_{oy} t - ½ g t²

        0 = y₀ + 0 - ½ g t²

        t² = 2 i / g

        t² = 2 4 / 9.81

        t² = 0.815

        t = 0.90 s

d) the horizontal distance traveled is

        x = v₀ₓ t

        x = 4 0.90

        x = 3.6 m

e) We can calculate the time to fall from I = 2 m

        y = y₀ + v_{oy} t - ½ g t²

        0 = y₀ + 0 - ½ g t²

        t² = 2 y₀i / g

        t² = 2 2 /9.81

        t² = 0.4077

        t = 0.639 s

Therefore, when making measurements, you should find readings around this value.