(^368) A Textbook of Engineering Mechanics
Notes : 1. In this case, the value of u is taken as negative due to upward motion.
- In this case, the distances in upward direction are taken as negative, while those in the
downward direction are taken as positive.
Example 17.9. A stone is thrown upwards with a velocity of 4.9 m/s from a bridge. If it falls
down in water after 2 s, then find the height of the bridge.
Solution. Given : Initial velocity (u) = – 4.9 m/s (Minus sign due to upwards) and time
taken (t) = 2 s.
We know that height of the bridge,
(^1122) (– 4 9 2) 9 8 (2) m
22
hut=+gt= ⋅×+×⋅×
= – 9.8 + 19.6 = 9.8 m Ans.
Example 17.10. A packet is dropped from a balloon which is going upwards with a velocity
12 m/s. Calculate the velocity of the packet after 2 seconds.
Solution. Given : Velocity of balloon when the packet is dropped (u) = – 12 m/s (Minus sign
due to upward motion) and time (t) = 2 s.
We know that when the packet is dropped its initial velocity (u) = – 12 m/s.
∴ Velocity of packet after 2 sec.
v = u + gt = – 12 + (9·8 × 2) = – 12 + 19·6 = 7·6 m/sAns.
Example 17.11. A body is dropped from the top of a tall building. If it takes 2.8 seconds in
falling on the ground, find the height of the building.
Solution. Given : Initial velocity (u) = 0 (because it is dropped) and time taken (t) = 2.8 s.
We know that height of the building
(^1122) (0 2.8) 9.8 (2.8)
22
sut=+gt=× +× × = 38.4 m Ans.
Example 17.12. A stone is dropped from the top of a building, which is 65 m high. With what
velocity will it hit the ground?
Solution. Given : Initial velocity (u) = 0 (because it is dropped) and height of the
building (s) = 65 m
Let v= Final velocity of the stone with which it will hit the ground.
We know that v^2 =u^2 + 2gs = (0)^2 + 2 × 9.8 × 65 = 1274
∴ v= 35.7 m/s Ans.
Example 17.13. A body is thrown vertically upwards with a velocity of 28 m/s. Find the
distance it will cover in 2 seconds.
Solution. Given : Initial velocity (u) = – 28 m/s (Minus sign due to upward motion) and time
(t) = 2 s.
We know that distance covered by the body,
(^1122) (– 28 2) 9.8 (2) m
22
sut=+gt= ×+× ×
= – 56 + 19.6 = – 36.4 m Ans.
....(Minus sign indicates that the body will cover the distance in upward direction)
Example 17.14. A bullet is fired vertically upwards with a velocity of 80 m/s. To what height
will the bullet rise above the point of projection?