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 when the hand is 2.10 above the ground. How long is the ball in the air before it hits the ground?

Answer 1

Ok, what's the equation of motion here? Let
u = initial velocity
s = initial position
g = acceleration due to gravity.

Then let y(t) be the position of the particle at time t, and v(t) its velocity at time t in the y direction.

We have that
\[ v(t) = u - gt \]
and
\[ y(t) = s + ut - \frac{1}{2}gt^2 \]

Now use that and the data given your equation to find the time t for which y(t) = 0; because that is when it reaches the ground.

Answer 2

thank you