Question

1. Before working this problem, review Conceptual Example 14. A pellet gun is fired straight downward from the edge of a cliff that is 12 m above the ground. The pellet strikes the ground with a speed of 28 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?

_______________________ m

2.  An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of +10 m/s and measures a time of 24.1 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet? (Indicate direction by the sign of the acceleration.)

_________________ m/s²

Answer 1

(1) Use the expression -

v^2 - u^2 = 2*a*s

Given that -

v = 28 m/s, u = ?, a = g = 9.8 m/s^2, s = 12 m

Put the values -

28^2 -u^2 = 2*9.8*12

=> u^2 = 548.8

=> u = 23.43 m/s

Now, when the pellet is fired upward -

u = - 23.43 m/s
When it reaches the maximum height v = 0.

s = v^2 -u^2 / 2a = -[-23.43]^2/ (2*9.8) = -28.0 m

Hence the pellet will go 28.0 m above the cliff. The negative sign shows the displacement in the upward direction.

(2) Suppose, a = acceleration due to gravity on the planet.

Initial velocity of the rock, u = +10.0 m/s

The rock returns in the hand of astronaut after time T = 24.1 s

At the highest point, velocity of the rock, v = 0

Use the expression -

v = u + a*t

=> 0 = u + a*t

=> t = -u/a

Total time of flight of the rock = T = 2*t = -2*(u/a)

So we have -

24.1 = -2*(10.0/a)

=> a = -24.1 / 20 = -1.205 m/s^2

So, the magnitude of the acceleration = 1.205 m/s^2. The negative sign shows that its direction is vertically downward direction.