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.
What conclusion can you draw about the velocity of a body from the displacement-time graph shown below :
A bus was moving with a speed of 54 km/h. On applying brakes it stopped in 8 seconds. Calculate the acceleration.
A car is travelling along the road at 8 ms-1. It accelerates at 1 ms-2 for a distance of 18 m. How fast is it then travelling ?
A car starting from rest acquires a velocity 180m s-1 in 0.05 h. Find the acceleration.
How can you find the following?
Acceleration from velocity – time graph.
Can you suggest a real-life example about the motion of a body from the following velocity – time graph?
How will the equations of motion for an object moving with a uniform velocity change?
Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of `"u"_1^2 : "u"_2^2` (Assume upward acceleration is –g and downward acceleration to be +g)
