(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey.
Hence, it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximum height, v = 0 m/s
Acceleration due to gravity, g = −9.8 ms−2
Using equation of motion, v = u + at, we have
0 = u + (−9.8 × 3)
⇒ u = 9.8 × 3 = 29.4 m/s
Hence, the ball was thrown upwards with a velocity of 29.4 m/s.
(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 m/s
Final velocity, v = 0 m/s
Acceleration due to gravity, g = −9.8 ms−2
Using the equation of motion,
= + 1/ 2 2
ℎ = 29.4 × 3 − 1/ 2 × 9.8 × 3 2 ⇒ ℎ = 44.1 m
Hence, the maximum height is 44.1 m.
(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0 m/s
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3 s = 1 s.
Using the equation of motion, = + 1/ 2 2
= 0 × 1 + 1/2 × 9.8 × 1 2 ⇒ = 4.9 m
Now, total height = 44.1 m
This means, the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds
