Question

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is released from the balloon. After it is released, for how long is the package in the air? The acceleration of gravity is 9.8 m/s 2 .

Answer 1

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$

So, the time the package was in the air is 1.97 seconds