A ball isthrown vertically upwards with a velocity of 49 m/s. Calculate
(i) themaximum height to which it rises.
(ii)thetotal time it takes to return to the surface of the earth.
A ball isthrown vertically upwards with a velocity of 49 m/s. Calculate
(i) themaximum height to which it rises.
(ii)thetotal time it takes to return to the surface of the earth.
(i)122.5 m (ii) 10 s
Accordingto the equation of motion under gravity:
v2− u2= 2 gs
Where,
u= Initial velocity of the ball
v= Final velocity of the ball
s= Height achieved by the ball
g= Acceleration due to gravity
Atmaximum height, final velocity of the ball is zero, i.e., v= 0
u= 49 m/s
Duringupward motion, g = − 9.8 m s−2
Leth be themaximum height attained by the ball.
Hence,
Lett be thetime taken by the ball to reach the height 122.5 m, then according tothe equation of motion:
v= u + gt
Weget,
But,
Timeof ascent = Time of descent
Therefore,total time taken by the ball to return = 5 + 5 = 10 s
(i)122.5 m (ii) 10 s
Accordingto the equation of motion under gravity:
v2− u2= 2 gs
Where,
u= Initial velocity of the ball
v= Final velocity of the ball
s= Height achieved by the ball
g= Acceleration due to gravity
Atmaximum height, final velocity of the ball is zero, i.e., v= 0
u= 49 m/s
Duringupward motion, g = − 9.8 m s−2
Leth be themaximum height attained by the ball.
Hence,
Lett be thetime taken by the ball to reach the height 122.5 m, then according tothe equation of motion:
v= u + gt
Weget,
But,
Timeof ascent = Time of descent
Therefore,total time taken by the ball to return = 5 + 5 = 10 s
