Quiz 1 Solutions

Physics 6A

1/15/03

Please read the questions carefully before attempting them. You will not be given any credit if you only write down the final answers. You must show your work.

1. A watermelon is dropped from a building starting from rest, (point A). After a time of $2.2s$ it passes by a window (point B). $0.5s$ after passing by the window, a large ``splat" is heard as it hits the ground (point C). Take the acceleration of gravity to be $9.8m/s^2$.

(a) (5 points) What was the velocity of the watermelon right before it hit the ground?

(b) (5 points) How high up is the top of the building, point A from the ground?

(c) (5 points) How high up is the window, point B from the ground?

Solutions

Define variables and coordinate system:

You can choose the origin where you want. I'll place $x=0$ at point A, and say the ball starts dropping at $t=0$, with $x$ increasing as the ball drops.

Call the position of point B in this coordinate system $x_B$ and call the time it passes by B, $t_B$. Same for $x_C$ and $t_C$.

What are the knowns?

We know $t_B = 2.2s$ and $t_C = t_B + 0.5s = 2.7s$. We know the initial velocity of the ball $v_0 = 0$ and it's initial position (for this choice of coordinate system) is $x_0 = 0$ and the acceleration of gravity is $a = +9.8m/s^2$.

Apply formulae

Now we can solve the problem because we can just use the formula for motion in 1d with constant acceleration.

(a) What was the velocity of the watermelon right before it hit the ground?

$v_C = v_0 + a t_C = 0 + 9.8 (m/s^2) 2.7 s = 26.46 m/s$ towards the ground.

(b) How high up is the top of the building, point A from the ground?

That'd be $x_C$, i.e. the distance between A and C, since $x_A=0$. But for constant acceleration we know $x_C = x_0 + v_0t + {1\over 2}at^2 = 0 + 0 {1\over 2} a t_C^2 = 0.5\times 9.8 )(m/s^2)\times (2.7s)^2 = 35.721 m$.

(c) (5 points) How high up is the window, point B from the ground?

First calculate $x_B$. It's just like part (b): $x_B = {1\over 2} a t_B^2 = 0.5\times 9.8 (m/s^2)\times (2.2s)^2 = 23.716 m$.

The height of the window off the ground is $x_C - x_B = 35.721 - 23.716 = 12 m.$

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