A. A particle is projected upward with velocity 60m/s.(g=10 m/s2 )
Find (a) average velocity of the particle in 8 seconds.
(b) average speed of the particle in 8 seconds.
Answers
Answer:
solved
Explanation:
60 - 10t = 0 , t = 6 s
max height H = 3600/20 = 180 m
displacement in 8s = 60x8 - 640/2 = 160 m
distance moved in 8s = 200 m
(a) average velocity of the particle in 8 seconds. = 160/8 = 20 m/s
(b) average speed of the particle in 8 seconds. = 200/8 = 25m/s
(a) Average velocity = 20m/s
(b) Average speed = 25m/s
Given:
Initial velocity 'u' = 60m/s
Acceleration due to gravity 'g' = 10m/s²
Time 't' = 8s
To find:
Average velocity
Average speed
Solution:
The maximum height the projected particle reaches
h = (v² - u²) / 2g , final velocity 'v' = 0
⇒ h = -u²/ 2g , since the particle is moving upwards u = -60m/s
Therefore, h = 60² / 20 = 3600/20
h = 180m
(a) Displacement in 8s
= ut + (gt²)/2 = -480 + 320
= -160m (negative sign indicates the upward motion)
Average velocity = Displacement/ Time = 160m / 8s
= 20m/s
(b) Time taken to reach the maximum height 180m = u/g
= (60m/s) / (10m/s²) = 6s
The distance travelled 2sec downward after reaching the maximum height 's' = gt²/2 = 20m
Distance travelled in 8s = h + s = 180m + 20m = 200m
Average speed = Distance / Time = 200m / 8s
= 25m/s
Therefore, the average velocity of the particle is 20m/s and the average speed of the particle is 25m/s.