Advertisements
Advertisements
Question
A packet is dropped from a stationary helicopter, hovering at a height ‘h’ from ground level, reaches the ground in 12s. Calculate
- the value of h
- final velocity of packet on reaching the ground. (Take g = 9.8 ms−2)
Advertisements
Solution
Height of the helicopter = h =?
Initial velocity = u = 0
Time = t = 12s
Acceleration = a = + g = + 9.8 ms−2
(1) S = ut + `1/2` at2
h = `0(12)+1/2(9.8)(12)^2`
h = 0 + 4.9 (144)
h = 705.6 m
(2) Let v = velocity of the packet on reaching the ground.
v = u + at
v = 0 + (9.8) 12
v = 117.6 ms−1
APPEARS IN
RELATED QUESTIONS
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h−1 in 10 minutes. Find its acceleration.
Give one example of a situation in which a body has a certain average speed but its average velocity is zero.
A freely falling object travels 4.9 m in 1st second, 14.7 m in 2 nd second, 24.5 m in 3rd second, and so on. This data shows that the motion of a freely falling object is a case of :
The slope of a speed-time graph gives:
A body starts from rest and acquires a velocity 10 m s-1 in 2 s. Find the acceleration.
A car starting from rest acquires a velocity 180m s-1 in 0.05 h. Find the acceleration.
The velocity-time graph of a body in motion is a straight line inclined to the time axis. The correct statement is ___________
How can you find the following?
Acceleration from velocity – time graph.
A stone thrown vertically upwards takes 3 s to attain maximum height. Calculate
- initial velocity of the stone
- maximum height attained by the stone. (Take g = 9.8 ms−2)
When will you say a body is at uniform acceleration?
