Aboy threw a small bundle toward his girlfriend on a balcony 10. meters above him. the bundle stopped rising in 1.5 seconds. how high did the bundle travel? was that high enough for her to catch it?

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)

Note: that the acceleration is negative as it is opposing the direction of motion (acting  down on the object, causing deceleration)

We will be using equations of motion for an accelerating object to solve first for initial velocity, in order to find the distance.

Rearranging the equation, and using the value 'g' as our acceleration, we get, u = v - at

u = 0 - (-9.8)(1.5) = 14.7m/s

Rearranging the equation, we get, s= v^(2) - u^(2) / 2a

s = 0^(2) - 14^(2) / 2(-9.8)                          

Note: the negative sign is not a part of the initial velocity, but a part of the equation

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!

answer: 4

explanation: 8

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