Answer to Question #164671 in Mechanics | Relativity for Godiya

Question

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?

Answer 1

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

$a)$

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

Negative sign indicates the displacement in downward direction.

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

$b)$

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

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

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

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

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

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

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

$c)$

At maximum height , $v=0$

Using equation $v^2-u^2=2as$ and substituting the values,

$0^2-13.1175^2=2(-9.8)s$

$\implies s=8.79\ m$

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