Question

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?

Answer 1

Answer: 4 s

Explanation:

Given

The ball leaves the hand of student with a speed of $u=19\ m/s$

When the hand is $h=2.5\ m$ above the ground

Using the equation of motion we can write

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

Substitute the values

$\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]$

Thus, the ball will take 4 s to hit the ground.