# 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 r

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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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Correct Answer - Maximum height is 122.5 m and total time si % s+5s =10s
Accrding to the equation of motion under gravity
v^(2)-u^(2)=2gs
where
u= intial velocity of the ball
v= final velocity of the ball
s= Height achieved by the ball
g= Acceleration due to gravity
At maximum height , final velocity of the ball is zero ,i.e., v=0
u=49 m/s
During upward , g=-9.8ms^(-2)
let h be the maximum height attained by the ball
Hence ,
0-49^(2)=2xx9.8xxh
implies h=49^(2)//(2xx9.8)
implies h=2401//19.6 =1225.5
Let t be the times taken by the ball to reach the height 122.5 m. then according to the equation of motion :
v=u+gt
We get
0=49+txx)-9.8)
9.8 t=49
t=49//9.8=5s
Also ,
time of ascent = Time of descent
therefore , total time taken by the ball to return =5+5=10 s