A ball thrown up vertically returns to the thrower after 3s. Find 3
a) the velocity with which it was thrown up,
b) the maximum height it reaches, and
c) its position after 2s.
Answers
(a) Time of ascent is equal to the time of descent. The ball takes a total of 3 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.
The ball returns to the ground after 6 seconds.
Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s
Let the velocity with which it is thrown up be u
(a). For upward motion,
v=u+at
∴ 0=u+(−10)×3
⟹u=30 m/s
(b). The maximum height reached by the ball
h=ut+
2
1
at
2
h=30×3+
2
1
(−10)×3
2
h=45 m
(c). After 3 second, it starts to fall down.
Let the distance by which it fall in 1 s be d
d=0+
2
1
at
′2
where t
′
=1 s
d=
2
1
×10×(1)
2
=5 m
∴ Its height above the ground, h
′
=45−5=40 m
Hence after 4 s, the ball is at a height of 40 m above the ground.
