1.

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

Answer»

Solution :ACCORDING to the equation of motion under gravity :
`v^(2)-u^(2)=2gs`
Where u= initial 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=o, u=49m/s. During upward motion, `g=9.8ms^(-2)`. LET be the maximum height attained by the ball. Henece, `0-49^(2)=2xx9.8xxh`
`H=(49xx49)/(2xx9.8)=122.5m`
Let be the time taken by the ball to REACH the height 122.5m, then according to the equation of motion.
v=u+gt
Weget `0=49+txx(-9.8)`
9.8t=49
`t=49//9.8=5s`
But time of ascent=time of descent. Therefore, total time taken by the ball to return `=5+5=10s`.


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