Question

A balloon rising vertically up with uniform velocity 15m/s releases a ball at a height of 100m. Calculate the time taken by the ball to hit the ground.(take g = 10 m/s^2).

Answer 1

Answer :

›»› The time taken by the ball to hit the ground are 6.2 sec.

Given :

• Initial velocity of balloon (u) = -15 m/s
• Releases a ball at a height (h) = 100 m
• Acceleration due to gravity (g) = 10 m/s²

To Calculate :

• Time taken by the ball (t) = ?

Calculation :

From second equation of motion

$\tt{: \implies h = ut + \dfrac{1}{2}g {t}^{2} }$

$\tt{: \implies 100 = - 15t + \dfrac{1}{2} \times 10 \times t}$

$\tt{: \implies 20 = - 3t + {t}^{2} }$

$\tt{: \implies {t}^{2} - 3t - 2 = 0}$

$\tt{: \implies t = \dfrac{3 \pm \sqrt{9 - 4( - 20)} }{2} }$

$\tt{: \implies t = \dfrac{3 \pm \sqrt{9 + 80} }{2} }$

$\tt{: \implies t = \dfrac{3 \pm \sqrt{89} }{2} }$

$\tt{: \implies t = \dfrac{3 + \sqrt{89} }{2} }$

$\tt{: \implies t = \dfrac{3 + 9.4}{2} }$

$\tt{: \implies t = \dfrac{12.4}{2} }$

$\frak{: \implies \underline{ \boxed{ \pink{ \frak{t = 6.2\: sec}}}}}$

║Hence, the time taken by the ball to hit the ground are 6.2 sec.║

Answer 2

Given :-

☄ Initial velocity of balloon, u = -15 m/s

☄ Releases a ball at a height, h = 100 m

☄ Acceleration due to gravity, g = 10m/s²

To find :-

Time taken by the ball to hit the ground

Solution :-

using second equation of motion .i.e.,

$\bf{\dashrightarrow h = ut + \dfrac{1}{2}g {t}^{2} } \\ \\ </p><p>\sf{\dashrightarrow 100 = - 15t + \dfrac{1}{2} \times 10 \times t} \\ \\ </p><p>\sf{\dashrightarrow 20 = - 3t + {t}^{2} } \\ \\ </p><p>\sf{\dashrightarrow {t}^{2} - 3t - 2 = 0} \\ \\ </p><p>\sf{\dashrightarrow t = \dfrac{3 \pm \sqrt{9 - 4( - 20)} }{2} } \\ \\ </p><p>\sf{\dashrightarrow t = \dfrac{3 \pm \sqrt{9 + 80} }{2} } \\ \\ </p><p>\sf{\dashrightarrow t = \dfrac{3 \pm \sqrt{89} }{2} } \\ \\ </p><p>\sf{\dashrightarrow t = \dfrac{3 + \sqrt{89} }{2} } \\ \\ </p><p>\sf{\dashrightarrow t = \dfrac{3 + 9.4}{2} } \\ \\ </p><p>\sf{\dashrightarrow t = \dfrac{12.4}{2} } \\ \\ </p><p>\frak{\dashrightarrow { \boxed{ \bf{t = 6.2\: sec}}}}$

Hence, the time taken by the ball to hit the ground is 6.2 sec