the function v (t) = -9.8t+5 gives the instantaneous velocity of an object thrown upward with an initial velocity of 5 m/sec. at what time t does the object start falling?

0.5 s (using 1 significan figure since the initial velocity is given with 1 significant figure)

Explanation:

Let's see what each term in the function v (t) = - 9.8t + 5 means:

Term: - 9.8 t

The variable t is the time

The coffecient - 9.8 (which is negative) means that the velocity is reduced 9.8 m/s each second.

That means that the motion is uniformly decelerated with acceleration - 9.8 m/s².

Term + 5

The term + 5, means that the object was launched with an upward velocity of 5 m/s.

Hence the equation tells that after the object was launched upward it starts to lose speed and will reach a maximum height when the velocity is equal to zero, from which it starts falling.

So, you can calculate when the object reachs it maximum height and starts falling by making the velocity, v (t), equal to zero and solving for t. This is how you do it:

v (t) = 0 0 = - 9.8t + 5 9.8t = 5 remember that the units of 9.8 is m/s² and of 5 is m/s

t = 5 m/s / 9.8 m/s² = 0.51 s

So, the object starts fallin at 0.51 s. Since, the velocity is reported with 1 significant digit you answer should also show 1 significant digit, which means t = 0.5 s.