A robot probe drops a camera off of the rim of a 239 m high cliff on Mars, where the free-fall acceleration is -3.7 m/s^2.
A. Find the time required for the camera to reach the ground.
B. Find the velocity with which it hits the ground.( shown work will be appreciated)
Sure! Let's solve the problem step by step.
A. Finding the time required for the camera to reach the ground:
Step 1: First, let's find the initial velocity (v₀) of the camera at the top of the cliff. Since the camera is dropped, its initial velocity is 0 m/s.
Step 2: Now, we can use the equation of motion to find the time (t) it takes for the camera to fall:
h = v₀ * t + (1/2) * g * t²
where:
h = height (239 m)
v₀ = initial velocity (0 m/s)
g = acceleration due to gravity (-3.7 m/s²)
Step 3: Plug in the values into the equation:
239 = 0 * t + (1/2) * (-3.7) * t²
Simplifying the equation further:
239 = -1.85t²
Step 4: Divide both sides of the equation by -1.85:
t² = -239 / -1.85
t² = 129.1892
Step 5: Take the square root of both sides to find t:
t = √129.1892
t ≈ 11.358 seconds
Therefore, the time required for the camera to reach the ground is approximately 11.358 seconds.
B. Finding the velocity with which it hits the ground:
Step 6: To find the velocity (v) with which the camera hits the ground, we can use the equation:
v = v₀ + g * t
Step 7: Substitute the values into the equation:
v = 0 + (-3.7) * 11.358
v ≈ -41.8766 m/s
Since the negative sign indicates the direction, the camera hits the ground with approximately 41.8766 m/s in the downward direction.
So, the velocity with which the camera hits the ground is approximately 41.8766 m/s (downward).
To find the time required for the camera to reach the ground, we can use the kinematic equation:
h = (1/2) * g * t^2 + v_0 * t + h_0
where:
h is the height (distance) - in this case, 239 m
g is the acceleration due to gravity - on Mars, it is -3.7 m/s^2 (negative because it acts downward)
t is the time we want to find
v_0 is the initial velocity (the probe was dropped, so v_0 = 0)
h_0 is the initial height (the camera was dropped from the cliff, so h_0 = 0)
Plugging in the values, the equation becomes:
239 = (1/2) * (-3.7) * t^2 + 0 * t + 0
Simplifying:
239 = -1.85 * t^2
Dividing both sides by -1.85:
-129.189 = t^2
Taking the square root of both sides (ignoring the negative square root, since time cannot be negative), we get:
t = sqrt(129.189) ≈ 11.36 s
So, the time required for the camera to reach the ground is approximately 11.36 seconds.
Next, to find the velocity with which it hits the ground, we can use the equation:
v = v_0 + g * t
where:
v is the final velocity
v_0 is the initial velocity (0 in this case)
g is the acceleration due to gravity (-3.7 m/s^2 on Mars)
t is the time (11.36 s)
Plugging in the values, the equation becomes:
v = 0 + (-3.7) * 11.36
Simplifying:
v = -42.332
So, the velocity with which it hits the ground is approximately -42.332 m/s. The negative sign indicates the velocity is downward.
A. 239 = ½ * 3.7 * t²
B. v = a t = -3.7 * t
... t is from part A