Question

A ball thrown vertically up words with a velocity of 20m/s from the top of a Multistoried.
The height of the point from where the ball is thrown is 25m from the ground.
(i) How high will the ball rise?
(ii) How long will it be before the ball hits the ground?​

Answer 1

Answer:

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

t=2s

s=ut+0.5at^2=20(2)+0.5(−10)(2)^2 =20m

Total distance for maximum height is 45 m

s=ut+0.5at^2

45=0+0.5(10)(t' )^2

t′ =3s

Total time= 3+2= 5s

Note - In this take g= 10m/s^-2

Answer 2

Answer :-

1) 45m

2) 5s

Explanation :-

Given :

Initial velocity of the ball,u = 20m/s

Final velocity of the ball,v = 0 (as it covers its maximum height)

Ball is thrown from a height,H of = 25m

To Find :

(i) How high will the ball rise?

(ii) How long will it be before the ball hits the ground?

Solution :

We know,

$\boxed{\sf{}v^2=u^2+2as}$

Here a is same as g (acceleration due to gravity) and s is same as height(h)

So,

$\sf{}:\implies v^2=u^2+2gh$

$\sf{}:\implies 0=25^2+ 2\times (-10)\times h$

$\sf{}:\implies -625=-20\times h$

$\sf{}:\implies \dfarc{-625}{-20}=h$

$\sf\therefore h =20$

Therefore,height from the ground 25 + 20 = 45m

According to the second law of equation,

$\boxed{\sf{}s=ut+\dfrac{1}{2}at^2}$

Put their values and find

Note:-

Here,net displacement, s = - 25 m (negative sign is taken as displacement is in the opposite direction of the initial velocity)

Let’s solve it step-by-step:-

$\sf{}:\implies -25=20\times t-\dfrac{1}{2}\times 10\times t^2$

$\sf{}:\implies 0=20\times t-\dfrac{1}{2}\times 10\times t^2+25$

$\sf{}:\implies 20\times t-\dfrac{1}{2}\times 10\times t^2+25=0$

$\sf{}:\implies 20\times t-5t^2+25=0$

$\sf{}:\implies -5t^2+20t+25=0$

$\sf{}:\implies 5(-t^2+4t+5)=0$

$\sf{}:\implies 5(-t^2+4t+5)\times \dfrac{1}{5}=0\times \dfrac{1}{5}$

$\sf{}:\implies -t^2+4t+5=0$

$\sf{}:\implies -(t^2-5t)-(t-5)=0$

$\sf{}:\implies -t(t-5)-1(t-5)=0$

$\sf{}:\implies (-t-1)(t-5)=0$

- t - 1 = 0

∴ t ≠ -1

t - 5 = 0

∴ t = 5

We know,time is always positive,so we not take -1

Therefore,time is equal to 5s