Question

43. A cricket ball is thrown up with a speed of 19.6 ms. The
maximum height it can reach is
(a) 9.8 m (b) 19.6 m (c) 29.4 m (d) 39.2 m​

Answer 1

Here we should use newton's second eqn. of motion 

that is= "V^2-U^2=2GH" 

So here u= 19.6m/s. and v=0m/s as after reaching a certain height v becomes 0.

now the eqn. changes to -u^2=2gh which can be written as h=-u^2/2g

And here even g is also neagative as the ball is thrown upwards

so, h= -384.16/2*-9.8

    h= 19.6m

Answer 2

Answer:

See.....For this type of questions you should know the equation of motions......

Here we will use third equation of motion that is V²=u²+2as......

Explanation:

As according to given problem we will modify our formula as

V²=u²-2gh.......where v is the final velocity, u is the initial velocity with which the ball was thrown, g is the gravitational acceleration.....we taken acceleration as gravitational acceleration because it is the question of free fall.....whenever you throw a ball it goes upward against the gravitational pull which acts downward due to which it gives an acceleration to the ball in downward direction as the ball is moving upward and acceleration is downward we take negative sign in our equation, which i showed in very first line.......

H represents the maximum height, also S in our formula represent height.....

Now after reaching maximum height velocity of ball will become zero as because the kinetic energy which you provided will get completely converted to potential energy as max height( always remember more potential energy less stable), so we can take our final velocity V as 0 and then our formula becomes

0=u²-2gh

or, u²=2gh

or..... h= u²/2g

u= 19.6 given in question and take g=9.8m/s² after putting values you will get answer as h=19.6m option (b)

I hope it will help you please feel free to ask any query.......