Question

Q.5 A stone is dropped from a balloon going up with a uniform velocity of 10 m/s. If the height of the balloon was 56.25 m when the stone was dropped, then the time after which the stone will strike the ground after it is dropped is [g = 10 m/s2

Answer 1

consider the motion of the stone in vertical direction

v₀ = initial velocity of stone just after it leaves the balloon = uniform velocity of balloon = 10 m/s

Y₀ = initial position of stone = height of balloon when stone leaves = 56.25 m

Y = final position of the stone when at ground = 0 m

a = acceleration of stone = g = acceleration due to gravity = - 10 m/s²

t = time of travel for stone before it strikes the ground

Using the kinematics equation

Y = Y₀ + v₀ t  + (0.5) a t²

inserting the above values in the equation

0 = 56.25 + 10 t  + (0.5) (- 10) t²

t = 4.5 sec