Question -
Answer -
(a)Time of ascent is equal to the time of descent. The ball takes a total of 6 sfor its upward and downward journey.
Hence,it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximumheight,┬аv┬а= 0
Acceleration due to gravity, g = тИТ9.8 m sтИТ2
Equation of motion,┬аv┬а=┬аu┬а+┬аgt┬аwillgive,
0 =┬аu┬а+ (тИТ9.8 ├Ч 3)
u┬а= 9.8 ├Ч 3 = 29.4 msтИТ1
Hence, the ball was thrown upwards with avelocity of 29.4 m sтИТ1.
(b) Let the maximum height attained by theball be┬аh.
Initial velocity during the upwardjourney,┬аu┬а= 29.4 m sтИТ1
Final velocity,┬аv┬а= 0
Acceleration due to gravity, g = тИТ9.8 m sтИТ2
Fromthe equation of motion,
(c)Ball attains the maximum height after 3 s. After attaining this height, it willstart falling downwards.
Inthis case,
Initial velocity,┬аu┬а= 0
Positionof the ball after 4 s of the throw is given by the distance travelled by itduring its downward journey in 4 s тИТ 3 s = 1 s.
Equationof motion,┬а
┬аwillgive,Totalheight = 44.1 m
Thismeans that the ball is 39.2 m (44.1 m тИТ 4.9 m) above the ground after 4seconds.