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 stationary position and moving with uniform acceleration attains a speed of 36 km per hour in 10 minutes. Find its acceleration.
A boy is sitting on a merry-go-round which is moving with a constant speed of 10 m s−1. This means that the boy is :
Define velocity. State its unit.
Differentiate between uniform acceleration and variable acceleration.
A bicycle initially moving with a velocity 5.0 m s-1 accelerates for 5 s at a rate of 2 m s-2. What will be its final velocity?
The velocity-time graph of a body in motion is a straight line inclined to the time axis. The correct statement is ___________
Derive the following equations for uniformly accelerated motion:
(i) v = u + at
(ii) `"S = ut" + 1/2 "at"^2`
(iii) v2 = u2 + 2aS
where the symbols have their usual meanings.
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)
The velocity-time graph (Fig. 8.5) shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds.
When will you say a body is at uniform acceleration?
