Question 18 A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4 s.
Class 9 - Science - Gravitation Page 144
Answers
t = 6/2 = 3 s.
v = 0 (at the maximum height )
a = - 9.8 m s-²
◆ a) Using, v = u + at, we get
0 = u - 9.8 × 3
or, u = 29.4 ms-¹
Thus, the velocity with which it was thrown up = 29.4ms-¹
◆ b) Using, 2aS = v² - u², we get
S = v²- u²/2a
= 0 - 29.4 × 29.4/- 2× 9.8
= 44.1 m
Thus, Maximum height it reaches = 44.1 m.
◆ c) Here, t = 4s. In 3 s, the ball reaches the maximum height and in 1 s it falls from the top.
Distance covered in 1 s from maximum height,
S = ut + 1/2at ²
= 0 + 1/2 × 9.8 × 1
= 4.9 m
Therefore, The ball will be 4.9 m below the top of the tower after 4 s.
(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
Acceleration due to gravity, g = −9.8 m s−2
Equation of motion, v = u + gt will give,
0 = u + (−9.8 × 3)
u = 9.8 × 3 = 29.4 ms− 1
Hence, the ball was thrown upwards with a velocity of 29.4 m s−1.
(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 m s−1
Final velocity, v = 0
Acceleration due to gravity, g = −9.8 m s−2
From the equation of motion,
(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
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.
Equation of motion, will give,
Total height = 44.1 m
This means that the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.