Question

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0 m/s when the hand is 1.90 m above the ground. You may want to review (Pages 49 - 51) . For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A How long is the ball in the air before it hits the ground

Answer 1

Answer:

t = 3.38 s

Explanation:

We have,

Initial speed of the ball that leaves the student's hand is 16 m/s

Initially, the hand is 1.90 m above the ground.

It is required to find the time for which the ball in the air before it hits the ground. We can use the equation of kinematics as :

$y_f=y_i+ut+\dfrac{1}{2}at^2$

Here, $y_f=-1.9\ m, y_i=0$ and a=-g

The equation become:

$-1.9=16t-\dfrac{1}{2}\times 9.8t^2$

After rearranging we get the above equation as :

$4.9t^2-16t-1.9=0$

It is a quadratic equation, we need to find the value of t. On solving the above equation, we get :

t = -0.115 s and t = 3.38 s (ignore t = -0.115 s )

So, the ball is in air for 3.38 seconds before it hits the ground.