Question

A student sees her physical science professor approaching on the sidewalk that runs by her dorm. She gets a water balloon and waits. When the professor is 2.0s from being directly under the window about 11m from the sidewalk, she drops the balloon. Finish the story.

d = 1/2 gt^2 1) I know time = t 2.0s 2)g = 9.8m/s^2    3) d=1/2 (9.8m/s^2)(2s)^2 4) I need help from here on out with step by step instructions please. I do not know what to multiply. Also, I am not using a scientific calculator.

Answer 1

Given that professor is 2.0 sec away from being directly under the window.

At that time balloon is dropped by student, So time taken by balloon to reach the ground will be:

Using 2nd kinematic equation in vertical direction:

h = U*t + (1/2)*a*t^2

Assuming downward direction as positive,

h = vertical displacement of balloon = 11 m

U = Initial speed of balloon = 0 m/s, since balloon was dropped

a = acceleration due to gravity = 9.8 m/s^2

So,

11 = 0*t + (1/2)*9.8*t^2

t = sqrt (2*11/9.8)

t = 1.50 sec = time taken by balloon to reach ground

Since given that professor is 2.0 sec away from directly being under the window, while balloon will reach at ground in 1.50 sec. So balloon will not hit the professor and it will also be in front of the eyes of professor, So student might get expelled from his school or it depends on the professor's mood.

Let me know if you've any query.