Question

A ball is thrown straight upward and rises to a maximum

heightof 16 m aboves its lauch point. At what height above
itslaunch point has the speed of the ball decreased to one-half of
itsinitial value?

Answer 1

Step-by-step explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

`v_f^2 - v_i^2 = 2 a d`

here we know that

`v_f = 0`

also we have

`a = -9.81 m/s^2`

so we have

`0 - v_i^2 = 2(-9.81)(16)`

`v_i = 17.72 m/s`

Now we need to find the height where its speed becomes half of initial value

so we have

`v_f = 0.5 v_i`

now we have

`v_f^2 - v_i^2 = 2 a d`

`(0.5v_i)^2 - v_i^2 = 2(-9.81)h`

`-0.75v_i^2 = -19.62 h`

`0.75(17.72)^2 = 19.62 h`

`h = 12 m`