Answer to Question #164671 in Mechanics | Relativity for Godiya

Question #164671

A ball is thrown along the vertical. The coordinate of the ball is observed as a function of time y(t). We choose y = 0 at the ground. The ball is at y = 13.2 m at time t = 0, and at y = 5.0 m at time t = 3.2 s. a) Find the displacement and average velocity of the ball between t = 0 and t = 3.2 s. b) Find the initial and final velocity of the ball. c) What is the height of the ball’s peak above ground? d) What is the average speed of the ball between t = 0 and t = 3.2 s?



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Expert's answer
2021-02-18T18:44:41-0500

The ball is thrown with a velocity uu from y=13.2 my=13.2\ m at t=0t=0 and it reaches to y=5 my=5\ m at t=3.2 mt=3.2\ m .

a)a)

Displacement of ball (s)=(513.2) m=8.2 m(s)= (5-13.2)\ m=-8.2\ m

Negative sign indicates the displacement in downward direction.

Average velocity of ball (vav)=8.23.2=2.5625 m(v_{av})=\frac{-8.2}{3.2}=-2.5625\ m


b)b)

a=9.8 m/sec2a=-9.8\ m/sec^2 (In downward direction )

Using equation s=ut+12at2s=ut+\frac{1}{2}at^2 and substituting the values,

8.2=u×3.2+12(9.8)(3.2)2-8.2=u\times 3.2 +\frac{1}{2}(-9.8)(3.2)^2

    u=+13.1175 m/sec\implies u=+13.1175\ m/sec which means ball is thrown in vertically upward direction.

Using v=u+atv=u+at and substituting the values,

v=13.1175+(9.8)(3.2)v=13.1175+(-9.8)(3.2)

    v=18.2425 m/sec\implies v=-18.2425\ m/sec which means ball is moving downwards.


c)c)

At maximum height , v=0v=0

Using equation v2u2=2asv^2-u^2=2as and substituting the values,

0213.11752=2(9.8)s0^2-13.1175^2=2(-9.8)s

    s=8.79 m\implies s=8.79\ m


d)d) Average speed =vav=2.5625 m/sec=2.5625 m/sec= |v_{av}|=|-2.5625|\ m/sec= 2.5625\ m/sec


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