Question

A hot - air balloon is rising vertically at a constant speed of 5 m.s^-1, when the binoculars of an occupant fall overboard. The binoculars will: 

Answer 1

Explanations

• The binocular too has the same velocity as the balloon: $\small 5 \,ms^{-1}$ upwards.
• As soon as it lose contact with the system of the balloon, it enters into a motion under the gravity, a vertical motion.
• As it has some velocity upwards it will travel some distance up & then begin to fall towards the ground.
• How much it rises can be calculated by the equation $\small v^2=u^2+2as$ applied to its vertical motion

$\qquad\qquad \begin{aligned} \small \uparrow 0^2 &=\small (5ms^{-1})^2+2(-9.8ms^{-2})h\\ \small h&=\small \frac{25}{2\times 9.8}\\ &=\small1.276\,m \end{aligned}$

During a time period of,

$\qquad\qquad \begin{aligned} \small \uparrow v&=\small u+at\\ \small 0&=\small 5+(-9.8)t\\ \small t&=\small 0.510\,s \end{aligned}$

• Any information regarding the motion can be calculated upon more data.