When a ball is thrown vertically upward it goes through a distance of 19.6 m find the initial velocity of ball and the time taken by it to rise the height point?

Question 38 Exercise 3E

Answer:

Given: maximum height, s= 20m, g= $10m/s^2$

(a)If ‘u’ is the initial velocity, velocity is zero at the highest point

We know that,

$\begin{array}{l} v^{2}-u^{2}=2 g s \\ 0-u^{2}=2(10)(20) \\ u^{2}=-400 m / s \\ u=20 m / s \end{array}$

Here, the negative sign indicates that the motion is against gravity.

(b) If v1 is the final velocity of the ball when it strikes the ground,

Velocity at the maximum height reached is zero which is equivalent to the initial velocity for the journey of the ball towards the ground.

Distance covered, s=20m;

We know from the equation of motion;

$\begin{array}{l} v^{2}-u^{2}=2 g s \\ v^{2}-0=2(10)(20) m / s \\ v^{2}=400 m / s \text { or } v=20 m / s \end{array}$

t=2 (20)/10

= 4s

Video transcript

hello everyone welcome to leader learning i'm good preet your science tutor today's question is a ball is thrown vertically upwards it goes to a height of 20 meters and then returns to the ground taking acceleration due to gravity g to be 10 meter per second square find a the initial velocity of the ball b the final velocity of the ball on reaching the ground and see the total time of the journey of the ball so let's start our answer here in the question maximum height is given that is s is equals to 20 meters the value of g is given as 10 meter per second square so for the a part of the question you have to calculate the value of u initial velocity u here the final velocity is zero at highest point okay at highest point the value of v is 0 now from the equation of motion v square minus u square is equals to 2g s put the values here v is 0 minus u square is equals to 2 into 10 into 20 that is u square is equals to minus of 400 meter per second you will get the value of u as minus 20 meter per second this is the required value of u and the negative sign shows that the motion is against the gravity now we have to calculate the value of v now this v can be calculated as we use the same formula v square minus u square is equals to 2 gs now put the initial velocity as 0 so v square minus 0 is equals to 2 into 10 into 20 so again you will get the value of v as 20 meter per second now in the last part of the question you have to calculate the value of t total time time can be given by the formula 2u by g okay now putting the values 2 the value of u is 20 upon g that is 10 you will get it as 4 second so these are the required answers for the question for more such videos please subscribe leader learning and for any doubts drop a comment thank you

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Text Solution

Solution : Here the ball is going up against the atteaction of earth so its velocity is decreasing continuously .In other words we can say that the ball is being retarded .Thus the acceleration in the ball is negative which means that the value of g is to used here with the neagtive sign. <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_PHY_IX_C03_S01_006_S01.png" width="80%"> Here,Initial velocity of ball u=? (To be calculated ) Final velocity of ball v= 0 (It stops ) Acceleration due to gravity `g= -9.8m//s^(2)` (Ball goes up ) Putting all these values in the formula : `v^(2)=u^(2)+2gh` we get `(0)^(2)=u^(2)+2xx(-9.8)xx19.6` `0=u^(2)-19.6xx19.6` `u^(2)=(19.6)^(2)` `u=19.6 m//s` Thus ,the intial velocity of the ball is 19.6 m/s which means that the ball has been thrown upwards with a velocity of 19.6 m/s . Let us now calculate the time taken by the ball to reach the highest point.Now we know the initial velocity ,the final velocity and the acceleration sue to gravity ,so the time taken can be calculated by using the equation : `v=u+gt` Final velocity v=0 (The ball stops ) Initial velocity u=19.6 m/s (Calculated above ) Acceleration due to gravity `g=-9.8 m//s^(2)` (Ball goes up) Time t=? (to be calculated ) So putting these values in the above equation we get `0=19.6+(-9.8) xxt` `0=19.6 - 9.8 t` 9.8 t = 19.6 `t=19.6/9.8` t=2s Thus the ball takes 2 seconds to reach the highest point of the ground .In other words ,the ball will take a total of 2+2=4 second to reach back to the thrower.