A boy throws rocks with an initial velocity of 12m/s [down] from a 20 m bridge into a river. Consider the river to be at a heigh
1 answer:
Answer:
Please find attached the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel
Explanation:
The given parameters are;
The initial velocity with which the boy throws the rock, u = 12 m/s
The direction in which he throws the rock = Down
The height from which the rock was thrown, h = 20 m
The height at which the river is located = 0 m
The kinematic equation of motion, of the rock can be given as follows;
h = u·t + 1/2·g·t²
Where;
t = The time of motion of the rock
g = The acceleration due to gravity = 9.8 m/s²
h = The height from which the rock is thrown = 20 m
Substituting the known values into given equation, we get;
20 = 12·t + 1/2·9.8·t² = 12·t + 4.9·t²
4.9·t² + 12·t - 20 = 0
t = (-12 ± √(12² - 4×4.9×(-20)))/(2 × 4.9) = (-12 ± √(242))/(9.8)
t ≈ -3.587 seconds or t ≈ 1.138 seconds
Attached please find the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel.
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