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 +16 m/s and measures a time of 23.2 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.)
Tr = 23.2/2 = 11.6 s. = Rise time.
Vf = Vo + g*Tr = 0.
16 + g*11.6 = 0.
11.6g = -16.
g = -1.38 m/s^2.
To determine the acceleration due to gravity on the distant planet, you can use the kinematic equation for vertical motion:
Δy = v0t + (1/2)gt^2
Where:
Δy is the displacement (change in height),
v0 is the initial velocity (in this case, +16 m/s),
t is the time taken (23.2 seconds), and
g is the acceleration due to gravity (what we want to find).
In this case, since the rock is thrown straight up and returns to the astronaut's hand, the displacement Δy will be zero.
Therefore, the equation simplifies to:
0 = (16 m/s)(23.2 s) + (1/2)g(23.2 s)^2
Let's solve for g:
0 = 371.2 m + (11.6 s^2)g
Rearranging the equation:
g = -371.2 m / (11.6 s^2)
Calculating this:
g ≈ -32 m/s^2
Hence, the magnitude of the acceleration due to gravity on this planet is approximately 32 m/s^2 (ignoring the negative sign for now). The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which means the gravity is pulling the rock downwards.
Therefore, the acceleration (magnitude and direction) due to gravity on this planet is approximately 32 m/s^2 in the downward direction.