Answer:
It will take 4 sec rock to comes its original point
Explanation:
It is given that the rock comes to its original point
So displacement S = 0 m
Initial velocity u = 19.6 m/sec
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
According to second equation of motion [tex]h=ut+\frac{1}{2}gt^2[/tex]
[tex]0=19.6\times t+\frac{1}{2}\times 9.8t^2[/tex]
[tex]19.6=4.9t[/tex]
t = 4 sec
Answer:
4seconds
Explanation:
The time interval that elapses between the rock’s being thrown and its return to the original launch point is known as its time of flight.
Time of flight is the time taken for an object to spend in the air after launch.
Time of flight is represented mathematically as
T = 2Usin(theta)/g where;
U is the initial velocity of the object = 19.6m/s
theta = angle of inclination between the object launched and the ground = 90° (since the body is thrown vertically upward)
g = acceleration due to gravity = 9.8m/s²
Substituting this values in the formula above we have;
T = 2(19.6)sin90°/9.8
T = 2(19.6)(1)/9.8
T = 4seconds
Therefore the time interval that elapses between the rock’s being thrown and its return to the original launch point is 2seconds
