Answer:
About 10.102 seconds
Step-by-step explanation:
Using the kinetics formula:
[tex]x = v_it + \frac{1}{2}at^2[/tex]
Substitute known values:
[tex]500=0t+\frac{1}{2}(9.8)t^2[/tex]
Solve for t:
[tex]1000=9.8t^2\\\frac{1000}{9.8}=t^2\\\sqrt{\frac{1000}{9.8}} =t\\10.102 \approx t[/tex]
Answer:
10.1 s
Step-by-step explanation:
Using equation of linear motion
h = ut + 1/2gt² where u = initial velocity = 0, the height from which the object is dropped = 500 m, g, acceleration due to gravity = 9.81m/s² and t is time in seconds
substitute the values into the equation
500 = 0t + 1/2gt²
500 × 2 = 9.81t²
1000 / 9.81 = t²
101.94 = t²
take square root of both side
√101.94 = t
t = 10.1 s
