CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A balloon starts rising from the ground with an acceleration of $$1.25 \;ms^{-2}$$, After $$8$$ seconds, a stone is released from the balloon, The stone will 
$$(g=10\;ms^{-2})$$


A
cover a distance of 40 m in reaching the ground.
loader
B
have a displaceent of 50 m
loader
C
reach the ground in 4 second
loader
D
begin to move down after being released
loader

Solution

The correct option is D reach the ground in 4 second
$$s = \dfrac {1}{2} at^2$$
$$a = 1.25\ m/s^2 \Rightarrow s = \dfrac {1}{2} \times 1.25 \times 8^2 = 40\ m$$
After $$8\ sec$$ speed of balloon:
$$v = u - gt$$
$$v = -(1.25) \times 8 = - 10\ m/s$$
Now, when stone is released distance to cover is $$40\ m$$.
$$v^2 = u^2 - 2gs$$
$$v^2 = (100)^2 + 2 \times 10 \times 40 \Rightarrow v = 30 m/s$$
$$v = u - gt \Rightarrow - 30 = +10 - 10t$$
$$t = 4\ s$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image