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 :
[tex]y_f=y_i+ut+\dfrac{1}{2}at^2[/tex]
Here, [tex]y_f=-1.9\ m, y_i=0[/tex] and a=-g
The equation become:
[tex]-1.9=16t-\dfrac{1}{2}\times 9.8t^2[/tex]
After rearranging we get the above equation as :
[tex]4.9t^2-16t-1.9=0[/tex]
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.
