Question

Calculate the displacement and velocity at 2.00 s for a ball thrown straight up with an initialvelocity of 15.0 m/s. Take the point of release to be y0= 10.0 m. How long does it take to hit the ground?Consider the origin of SOC at the ground and Oy is upward.

Answer 1

The displacement, velocity and time is 20.4 m, -4.52 m/s and 3.624 sec.

Step-by-step explanation:

Given that,

Initial velocity =15.0 m/s

Height = 10 m

We need to calculate the displacement at 2 sec

Using equation of motion

`s=ut-(1)/(2)gt^2+s_(0)`

Put the value into the formula

`s=15t-(1)/(2)*9.8* t^2+10`

At 2 sec,

`s=15*2-(1)/(2)*9.8*4+10`

`s=20.4\ m`

We need to calculate the time

Using equation of motion again

`s=15t-(1)/(2)*9.8* t^2+10`

At ground s = 0,

`15t-(1)/(2)*9.8* t^2+10=0`

`4.9t^2-15t-10=0`

`t = 3.624\ sec`

Neglect the negative term because the time is not negative

We need to calculate the velocity

Using formula of velocity

`v=u-gt`

`v=15-9.8*3.624`

`v=-4.52\ m/s`

Hence, The displacement, velocity and time is 20.4 m, -4.52 m/s and 3.624 sec.