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

Answers

  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.

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