Hi!
We know that the boy must have thrown the bundle with some initial velocity 'v', and that the bundle stops after 1.5 seconds at a height 's' with a final velocity 'u' -which would be 0.
Our known values thus are:
Time (t) = 1.5 s
Final velocity (u) = 0 m/s
Gravitational acceleration (g) = -9.8 m/s^(2)
<em>Note: that the acceleration is negative as it is opposing the direction of motion (acting down on the object, causing deceleration)</em>
We will be using <em>equations of motion for an accelerating object</em> to solve first for initial velocity, in order to find the distance.
<h3>Initial Velocity using <u>v = u + at</u></h3>
Rearranging the equation, and using the value 'g' as our acceleration, we get,<em> u = v - at</em>
u = 0 - (-9.8)(1.5) = 14.7m/s
<h3>Distance Travelled using <u>
v^2 = u^2 + 2as</u></h3>
Rearranging the equation, we get, <em>s= v^(2) - u^(2) / 2a</em>
s = 0^(2) - 14^(2) / 2(-9.8)
<em>Note: the negative sign is not a part of the initial velocity, but a part of the equation </em>
s = - 216.09 / -19.6 = 11.025 m
The distance traveled by the bundle is hence, ~ 11 meters, which is high enough for the girl to catch.
Hope this helps!
