Question

A balloon rising vertically up with uniform velocity 15m/s releases a ball at height of 100m.calculate the time taken by the ball to hit the ground.take g _9.8m/ssqu

Answer 1

Answer:

Taking vertically upward direction to be positive

Initial velocity of the ball(u) = -15 m/s

Displacement of the ball(S) = 100 m

Acceleration = g = 10 m/s2

Let the time taken by the ball to reach the ground be t

Applying S=ut+12at2

100=−15t+1210×t2or, 20=−3t+t2t2−3t−20=0t=3±9−4(−20)√2=3±89√2t=3+89√2=6.21 sec

Answer 2

Answer:

Time taken by ball to hit the ground is $1.589$ seconds.

Explanation:

Let us take the vertically upward direction to be positive.

Initial velocity of the ball is given as $u = -15 m/s$

Displacement of the ball $S = 100 m$

Acceleration due to gravity $g=9.8 m/s^2$

Now, let the time taken by the ball to reach the ground be taken as $t$

We know the formula for displacement $S=ut+\frac{1}{2} at^2$

Substituting the values,

$100=-15t+\frac{1}{2}*9.8t^2$

$100=-15t+4.9t^2$

Taking every value to RHS : $49t^2-15t-100=0$

We can see that it is a quadratic equation in terms of $t$

Solving for this : $t=$ ± $\frac{-b+/-\sqrt{b^2-4ac}}{2a}$

$t=1.589$ or $t=-1.283$

Since $t$ cannot be a negative value, the correct answer is $t=1.589$ seconds.