Question

A rock is thrown straight up with an initial velocity of 19.6 m/s. What time interval elapses between the rock’s being thrown and its return to the original launch point? (Acceleration due to gravity is 9.80 m/s2 .)

Answer 1

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 $g=9.8m/sec^2$

According to second equation of motion $h=ut+\frac{1}{2}gt^2$

$0=19.6\times t+\frac{1}{2}\times 9.8t^2$

$19.6=4.9t$

t = 4 sec

Answer 2

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