ches the ground vd compare to the speed of the rock when it is thrown upward vu?
____vd > vu____vd = vu____vd < vu
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Sage Luong
2 years ago
Posted 2 years ago. Direct link to Sage Luong’s post “*note: the _ symbol refers to a subscript. So, v_f…”
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Good Answer
*note: the _ symbol refers to a subscript. So, v_f refers to the final velocity.
Sorry, but I believe the acceleration is -5m/(s^2).
x_f = x_i + v_i(t) + 0.5at^2
0m = 100m + (15m/s)(10s) + 0.5a(10s)^2
-100m = 150m + 50s^2a
-250m = 50(s^2)a
a = -5m/s^2
Now to the answer. The velocity as the rock reaches the ground will be greater than the initial velocity. Just try substituting values to test this out.
if object is dropped from height then it will perform dynamic motion and it is under free fall condition
Answer:
ches the ground vd compare to the speed of the rock when it is thrown upward vu?
____vd > vu____vd = vu____vd < vu
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Reply to Aimen’s post “An astronaut holds a rock at 100m above the surfac…”Button opens signup modal
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Comment on Aimen’s post “An astronaut holds a rock at 100m above the surfac…”Button opens signup modal
duskpin sapling style avatar for user Sage Luong
Sage Luong
2 years ago
Posted 2 years ago. Direct link to Sage Luong’s post “*note: the _ symbol refers to a subscript. So, v_f…”
Great Answer
Good Answer
*note: the _ symbol refers to a subscript. So, v_f refers to the final velocity.
Sorry, but I believe the acceleration is -5m/(s^2).
x_f = x_i + v_i(t) + 0.5at^2
0m = 100m + (15m/s)(10s) + 0.5a(10s)^2
-100m = 150m + 50s^2a
-250m = 50(s^2)a
a = -5m/s^2
Now to the answer. The velocity as the rock reaches the ground will be greater than the initial velocity. Just try substituting values to test this out.
v_f = v_i + at
v_f = 15m/s – 5m/s^2(10s)
v_f = 15m/s – 50m/s
v_f = -35m/s
speed_f = |v_f|
speed_f = |-35m/s|
speed_f = 35m/s
Therefore, speed_f > speed_i.
Explanation:
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