Question

A ball is tossed up into the air with an initial speed of 5.0. How long does it take to return to the person's hand

Answer 1

Answer:

Explanation:

$y(t) = 5t-5t^2$

When y(t) = 0, the ball is on the hand.

$0=5t-5t^2\\0=5t(1-t)\\t=0,1$

It takes 1 second.

Answer 2

There are several different possibilities.

==>  If the 5.0 means 5 miles per hour, that's 2.24 meters per second, up.

The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (2.24/9.8) = 0.229 second.  Then, after is starts to fall, it takes the same amount of time to the person's hand.

Total time = 0.457 second.

==>  If the 5.0 means 5 meters per second, up . . .

The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (5.0/9.8) = 0.51 second.  Then, after is starts to fall, it takes the same amount of time to the person's hand.

Total time = 1.02 second.

==>  If the 5.0 means 5 km/minute, that's about 83.33 meters per second, up.

The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (83.33/9.8) = 8.503 seconds.  Then, after is starts to fall, it takes the same amount of time to the person's hand.

Total time = 17.01 seconds.

==>  If the 5.0 means 5 furlongs per fortnight, that's about 0.00083 meters per second, up.

The acceleration of gravity is 9.8 m/s down, so the ball stops, turns around, and starts falling in (0.00083/9.8) = 0.000085 second.  Then, after is starts to fall, it takes the same amount of time to the person's hand.

Total time = 0.00017 second.

This is why all of your numbers always need their units.