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DENIUS [597]
3 years ago
8

A boy threw a small bundle toward his girlfriend on a balcony 10. meters above him. The bundle stopped rising in 1.5 seconds. Ho

w high did the bundle travel? Was that high enough for her to catch it?
Biology
1 answer:
Alisiya [41]3 years ago
6 0

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!


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