Question

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

Answer 1

(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, $g=-9.8m{s}^{-2}$
From the first equation of motion,
v = u + gt
$0=u+(-9.8\times 3)$
$u=9.8\times 3=29.4m{s}^{-1}$
Hence, the ball was thrown upwards with a velocity of $29.4m{s}^{-1}$

(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, $u=29.4m{s}^{-1}$
Final velocity, v = 0
Acceleration due to gravity, $g=-9.8m{s}^{-2}$
From the second equation of motion,
$s=ut+\frac{1}{2}g{t}^2$
$h=29.4\times 3+\frac{1}{2}\times -9.8\times (3{)}^2=44.1m$

(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
Distance travelled by it during its downward journey in remaining 1 s is given by
$s=0\times t+\frac{1}{2}\times 9.8\times 1^2=4.9m$
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.