Question

A ball is thrown vertically upward.It falls back to the ground (same spot) after 2 second.

Find the height reached by the ball. (g=9.8 m/s

2

)​

Answer 1

Answer:

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.

(Ball goes up ) Putting all these values in the formula :

we get

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 . <br> 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 :

Final velocity v=0 (The ball stops ) Initial velocity u=19.6 m/s (Calculated above ) Acceleration due to gravity

(Ball goes up) Time t=? (to be calculated ) So putting these values in the above equation we get

9.8 t = 19.6

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.