If a ball is thrown upward on Mars with an initial velocity of 10.0 m/s and lands 1.50 m below its launch point 5.54 seconds after it is thrown , what is the local acceleration due to gravity one Mars .... Plzzzzzzzzz answer it's really important I have to submit it tomorrow ... I know the answer should be 3.71 m/s^2 just please tell me how can I solve it ( show work ) . I tried every way but I couldn't get anything with that answer ... So plz help and thank you
Give me the answer pleaseeeeeeeeee i rally have to know it I just have this one question :( ... I know the answer but I don't know how to solve it . 😔
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To solve this problem, we can use the kinematic equation that relates displacement (d), initial velocity (v0), time (t), and acceleration (a):
d = v0 * t + 0.5 * a * t^2
Given:
Initial velocity (v0) = 10.0 m/s
Displacement (d) = -1.50 m (negative because the ball lands below its launch point)
Time (t) = 5.54 s
We need to solve for the acceleration (a). Rearranging the equation, we get:
a = (d - v0 * t) / (0.5 * t^2)
Substituting the given values:
a = (-1.50 m - 10.0 m/s * 5.54 s) / (0.5 * (5.54 s)^2)
a = (-1.50 m - 55.4 m) / (0.5 * 30.7316 s^2)
a = (-56.9 m) / (15.3658 s^2)
a ≈ -3.71 m/s^2 (rounded to two decimal places)
Therefore, the local acceleration due to gravity on Mars is approximately 3.71 m/s^2.