Question

A ball is thrown vertically upwards at an initial speed of 23 ms". What is the displacement of the ball 3sec after it was thrown?

Answer 1

$\huge\underline\red{SOLUTION}$

$\underline\red{GIVEN}$

Speed=23ms

$\underline\red{ANSWER}$

Let's take u=23m/s

First calculate the time taken to reach the highest point. At that point velocity is 0 . So by v=u+at, 0=u-gt( final velocity at the highest point is zero and the acceleration due to gravity is acting downwards so negative sign, taking down

So,u=gt

♠t= u/g=23/9.8=2.35s

♠So in 2.35s the ball reached the highest point.

♠Now let's find the displacement in 2.35s

♠By v2-u2=2as, a=-g( acting downwards) , v=0

♠We get, 0-u2=-2gs

♠u2=2gs

♠s=u2/2g

♠s= 23×23/2×9.8=26.98 m

Now it will come down. So let's consider initial velocity 0. We have to find the displacement in 3 sec. We know that 26.98 m is covered in 2.35s. So we need to find the distance covered in

♠3–2.35=0.65s. So t=0.65. By s =ut+0.5at^2.

♠u=0, a=g(considering downwards positive)

♠s=0×0.65+0.5×9.8×0.65×0.65

♠s=2.07

Now since we need to find the displacement in 3s you need to observe that the answer will be 26.98–2.07=24.91m. The ball went up to 26.98m and came down 2.07m (all in 3 sec) so the total distance covered will be 26.98+2.07 but if we need to find displacement we need to subtract.