# A ball is thrown vertically upwards with a velocity of 49 m/s Calulate: (i) the maximum height to which it rises . (ii) the total time it takes to ret

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A ball is thrown vertically upwards with a velocity of 49 m/s Calulate:
(i) the maximum height to which it rises .
(ii) the total time it takes to return to the surface of the earth .

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Please note that here the ball is going up against the gravity so the value of acceleration due t gravity g is t be taken as negative ( with a minus sign)
Here , Initial velocity of ball u=49 m/s
Final velocity of ball v=0 (The ball stops at top)
Acceleration due to gravity g=-9.8 m//s^(2) (The ball goes up)
And Height ,h=? (To be calculated )
Now ,Putting all these values in the formual
v^(2)=u^(2)+2gh
(0)^(2)=(49)^(2)+2xx(-9.8)xxh
0=2401-19.6 h
19.6 h= 2401
h=(2401)/(19.6)
h=122.5 m
Thus the maximum height to which the ball rises is 122.5 metres.
(ii)We will first calculate the time taken by the ball to reach the highest point by using the formula :
v=u+gt Here Final velocity v=0 (the ball stops at top )
Initial velocity u= 49 m/s
Acceleration due to gravity g=-98. m//s^2 (The ball goes up )
And so putting these value in the above formula ,we get :
0=49 +(-9.8)xxt
0=49 - 9.8 t
9.8 t =49
t=(49)/(9.8)
t= 5 s
Thus the ball takes 5 seconds to reach the highest point of its upwards journey .Please note that ball will take an equal time ,that ,is 5 seconds to reach return to the surface of the earth .In ohter words the ball will take a total time of 5+5 =10 seconds to return to the surface of the earth .