A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the
hand is 2.50 m above the ground.
Required:
How long is the ball in the air before it hits the ground?
1 answer:
Answer: 4 s
Explanation:
Given
The ball leaves the hand of student with a speed of 
When the hand is 
above the ground
Using the equation of motion we can write
Substitute the values
![\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]](https://tex.z-dn.net/?f=%5CRightarrow%202.5%3D-19t%2B0.5%5Ctimes%209.8t%5E2%5C%5C%5CRightarrow%204.9t%5E2-19t-2.5%3D0%5C%5C%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B19%5Cpm%20%5Csqrt%7B%28-19%29%5E2-4%5Ctimes%204.9%5Ctimes%20%28-2.5%29%7D%7D%7B2%5Ctimes%2019%7D%5C%5C%5CRightarrow%20t%3D4.0049%5Cquad%20%5B%5Ctext%7BNeglecting%20the%20negative%20value%20of%20%7Dt%5D)
Thus, the ball will take 4 s to hit the ground.
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