2.
A particle is projected vertically up from the top of a tower with velocity 10 m/s. It reaches the ground
in 5s. Find-
(a) Height of tower.
(b) Striking velocity of particle at ground
(c) Distance traversed by particle.
(d) Average speed & average velocity of particle.
is released from the
Answers
Answer:
The height or distance traveled by the particle is 83.5 m.
Explanation:
Given that,
Velocity v = 10 m/s
Time t = 5 sec
We need to calculate the distance from the top of a tower
Using third equation of motion
v^2=u^2+2gsv
2
=u
2
+2gs
Where, v= final velocity
u = initial velocity
g = acceleration due to gravity
s = distance
Put the value in the equation
0=10^2-2\times9.8\times s'0=10
2
−2×9.8×s
′
s'=\dfrac{100}{2\times9.8}s
′
=
2×9.8
100
s' =5.1\ ms
′
=5.1 m
For time,
The time at maximum height,
Using first equation of motion
v = u+gtv=u+gt
0=10-9.8\times t'0=10−9.8×t
′
t'=\dfrac{10}{9.8}t
′
=
9.8
10
t' =1.0\ sect
′
=1.0 sec
The maximum time is
t''=t-t't
′′
=t−t
′
t''=5-1=4\ sect
′′
=5−1=4 sec
Now, the particle reached at the ground from the maximum time in 4 sec.
For the distance traveled by the particle is
Using second equation of motion
s=ut+\dfrac{1}{2}gt^2s=ut+
Put the value in the equation
s''=0+\dfrac{1}{2}\times9.8\times4\times4s
′
×9.8×4×4
s''=78.4\ ms
′′
=78.4 m
The total distance is
s = s'+s''s=s
′
+s
′′
s=5.1+78.4=83.5\ ms=5.1+78.4=83.5 m
Hence, The height or distance traveled by the particle is 83.5 m.