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

Answer:

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

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

$s=ut-\dfrac{1}{2}gt^2+s_{0}$

Put the value into the formula

$s=15t-\dfrac{1}{2}\times9.8\times t^2+10$

At 2 sec,

$s=15\times2-\dfrac{1}{2}\times9.8\times4+10$

$s=20.4\ m$

We need to calculate the time

Using equation of motion again

$s=15t-\dfrac{1}{2}\times9.8\times t^2+10$

At ground s = 0,

$15t-\dfrac{1}{2}\times9.8\times t^2+10=0$

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

$t = 3.624\ sec$

Neglect the negative term because the time is not negative

We need to calculate the velocity

Using formula of velocity

$v=u-gt$

$v=15-9.8\times3.624$

$v=-4.52\ m/s$

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