Question

5) A bowling ball is dropped on the Jupiter from a height of 3 meters with an initial velocity of 0.4

Answer 1

Answer:

0.48 s

Explanation:

The vertical position of the ball at time t is given by:

$y(t) = h - ut - \frac{1}{2}gt^2$

where

h = 3 m is the initial height

u = 0.4 m/s is the initial velocity

g = 24.5 m/s^2 is the acceleration of gravity on the surface of Jupiter

The negative signs are due to the fact that the direction of the initial velocity and of the acceleration are downward

The ball reaches the surface at time t such that y(t)=0, so:

$0 = h-ut-\frac{1}{2}gt^2$

Substituting values,

$0=3-0.4t-12.25t^2$

Solving the second-order equation, we find:

$t=\frac{0.4\pm \sqrt{0.4^2-4(3)(-12.25)}}{2(-12.25)}$

which gives two solutions:

t = 0.48 s

t = -0.51 s

We discard the second one since it is negative, so the ball reaches the surface after 0.48 s.