Respuesta :
Answer:
Time taken to hit the floor, t = 0.61 s
Option a.
Given:
t = 0.50 s
Solution:
Initial velocity of the ball when dropped, u = 0 m/s
Now, using second equation of motion when acceleration due to gravity acts:
[tex]h = ut + \frac{1}{2}gt^{2}[/tex]
[tex]h = 0 + \frac{1}{2}\times 9.8\times 0.5^{2}[/tex]
h = 1.225 m
Now, when the lift moves in the upward direction with a constant speed of 1.0 m/s, u = - 1.0 m/s
Time taken by the ball to hit the floor travelling a distance h = 1.225 m is given by:
[tex]h = ut + \frac{1}{2}gt^{2}[/tex]
Now,
[tex]1.225 = - 1.0t + \frac{1}{2}\times 9.8\times t^{2}[/tex]
[tex]4.9t^{2} -t -1.225 = 0[/tex]
Solving the above quadratic equation, we get:
t = 0.612 s = 0.61 s
