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Question

. Astronauts on a distant planet throw a rock straight upward and record its motion with a video camera. After digitizing their video, they produce the graph of height, y, versus time, t, shown below. 

a. What is the acceleration of gravity on this planet? 

b. What was the initial velocity of the rock?

Answer 1

Let

y(t) be the height of the rock at a time t,

v(t) be the speed of the rock at a time t,

v(0) be the initial speed,

t1 = 4s be the time it takes the rock to reach the maximum height,

y(t1) = 30m be the maximum height,

a (to be found) be the gravitational acceleration

The trajectory of the rock is a combination of two movements:

- upward movement at constant speed of v(0), which is what the rock would do in the absence of gravity

- downward fall with acceleration a, which is what the rock would do in the absence of any initial speed.

The resulting speed is then

v(t) = v(0) - at

The resulting height is then

y(t) = v(0)t - at²/2

At maximum height at time t1,

v(t1) = 0,

y(t1) = 30m

That gives two equations

v(0) - 4a = 0

4v(0) - 16a/2 = 30

Whose solution is

a = 3.75 m/s²

v(0) = 15 m/s