Question

A balloon is rising vertically up with a velocity of $29m{s^{ - 1}}$ . A stone is dropped from it and it reaches the ground in 10s. The height of the balloon when the stone was dropped from it is: (g $ = 9.8m{s^{ - 2}}$ ).
A. $400m$
B. $150m$
C. $100m$
D. $200m$

Answer 1

Hint:Students should use motions equation with proper signs. Take the upper direction with a positive sign. The initial velocity will be the same with that of the balloon since the stone was thrown from that only.

Complete step by step answer:
Let the height of the tower be $h$
Now we take the position of the stone at t=0 as the origin and thus make the sign convention with the reference to it
We use Motion’s second equation for this question which gives that
$s = ut + \dfrac{1}{2}a{t^2}$
By using sign convention we get that
$
s = - h \\
\Rightarrow a = - g = - 9.8m{s^{ - 2}} \\
\Rightarrow u = 29m{s^{ - 1}} \\ $
Thus we put all the values of the variables and finally get the value of height
$
- h = 29 \times 10 - \dfrac{1}{2} \times 9.8 \times 10 \times 10 \\
\Rightarrow - h = 290 - 490 \\
\Rightarrow - h = - 200 \\
\therefore h = 200m \\ $
Thus the height of the tower is $200m$.Hence option (D) is the correct answer.

Additional Information:
Using these three we can solve any problem in one dimension. To solve two dimension problems, we resolve the velocities and accelerations in both the x and y axis and then apply Newton's law of motion separately.

Note:Students should be specifically careful in applying sign convention. Remember that we use the mathematics coordinate system to get the direction of positive and negative quantities. However we can make any point as origin and according give the signs of the quantities according to our will.