Question

In major league baseball, the pitcher's mound is 60 feet from the batter. If a pitcher throws a 85 mph fastball, how much time elapses from when the ball leaves the pitcher's hand until the ball reaches the batter? Express your answer to two significant figures and include the appropriate units.

Answer 1

The given from your problem are the following:
V = 85mph (This is miles per hour)
d = 60 feet

If you notice the units do not match. Before we can do anything else, we need to make the figures match. 

In this case, we will convert 85miles per hour to feet per hour.  There are 5,280 feet in 1 mile. 
$\frac{85miles }{hr}$ x $\frac{5,280feet}{1miles} = \frac{448,800feet}{hr}$

But wait! If you think about the scenario, you are looking for how long it will take for the ball to reach the bat. The most applicable unit of time to use here is second. It would be very hard to really measure a short and instantaneous event in hours. So we convert it into feet per second: 

There are 3,600 seconds in 1 hour.

$\frac{448,800feet }{hour}$ x $\frac{1hour}{3,600seconds} = \frac{448,800feet}{3,600 seconds} = 124.67ft/s$

So now we have our new given as:

v = 124.67ft/s
d = 60 ft

The formula for time can be derived from the formula from velocity, which is:
$velocity = \frac{distance}{time}$

The formula of time will then be:
$time= \frac{distance}{velocity}$

All you need to do is plug in what you know and solve for what you don't know. 

$time= \frac{60feet}{124.67ft/s}$

$time= 0.48s$

The answer then is 0.48s.

If you want this in hours, just divide the value in seconds by 3,600. The answer would then be 0.00013hr. (See how small it is? This is why seconds would be a more appropriate measure.)