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.

Answer

Maximum height is 122.5 m and total time si % s+5s =10s

Explanation

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`