# A rock is throw straight up. The thrower catches the rock in her hand 4.70 s later. a) What was the initial speed with which she threw the rock?  b) How high did the rock go?  Note: Consider...

A rock is throw straight up. The thrower catches the rock in her hand 4.70 s later.

a) What was the initial speed with which she threw the rock?

b) How high did the rock go?

Note: Consider gravitational force.

Expert Answers
electreto05 | Certified Educator

First, let's use the speed equation for any time of the vertical movement:

v = v0 ± gt

Where:

v;  is the speed for each moment.

v0;  is the initial velocity of the rock.

g = 9.8 m/s^2; is the acceleration of gravity.

For the calculation of the initial velocity of the rock; we analyze only the upward movement. In this case the final velocity is zero and the rising time is equal to half the total time of the movement. Also, in the equation, we take the negative sign.

Then we have:

v = 0;  tr = t/2 = 2.35 s

0 = v0 – g(tr)

v0 = g(tr) = (9.8)(2.35) = 23.03 m/s

To calculate how high the rock arrives, we apply the equation of distance for vertical movement, considering that the time is half the total time:

h = v0t - g(t^2)/2

h = (23.03)(2.35) – (9.8)(5.52)/2

h = 54.12 – 27.05

h = 27.07 m

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