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 when the hand is 1.80 above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Answer 1

There are mistakes in the question as the unit of speed and height is not mention here.The correct question is here

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0m/s when the hand is 1.80m above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Answer:

t=3.37s

Step-by-step explanation:

Given Data

As we have taken hand at origin and positive upward

So given data are

`y_(i)=0m\\y_(f)=-1.80m\\v_(i)=16.0m/s\\a=g=9.8m/s^(2)`

To find

time taken by the ball before it hits the ground

Solution

By using the common kinematic equation

`y_(f)=y_(i)+v_(i)t+0.5at^(2)`

Put the given values and find for t

So

`-1.80=0+16.0t+(0.5*(-9.8)t^(2) )\\-1.80=16.0t-4.9t^(2)\\ 4.9t^(2)-16.0t-1.80=0`

Apply quadratic formula to solve for t

`t=\frac{-(-16.0)+\sqrt{(-16)^(2)+4(4.9)(-1.80)} }{2(4.9)}\\ t=3.37s`