Question

An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s².

Answer 1

From newton’s third equation of motion,

V² = u² + 2as

Where:

V = Final velocity;

U = initial velocity;

T = time taken;

S = distance travelled;

A = acceleration;

According to our question,

The figure below illustrates the situation given in the question

V = 0 (Velocity at maximum height is zero);

S = 500 m;

A = −10 m/s² (because when the object is going up, the acceleration due to gravity will be acting downwards to make the object fall). Hence by sign convention direction of motion and acceleration is opposite (therefore, a is negative);
Putting the above values we get

0² = u² + (2 x (−10) x 500)

0 = u² − 10000

u² = 10000

⇒ u = `sqrt10000` = 100

Therefore, the initial velocity is 100 m/s.

From the first equation of motion,
v = u + at

Where symbols have there usual meanings as above;

v = 0 (velocity at maximum height is zero);

u = initial velocity = 100 m/s;

a = −10 m/s² (because when the object is going up, the acceleration due to gravity will be acting downwards to make the object fall). Hence by sign convention direction of motion and acceleration is opposite (therefore, a is negative);
Putting the values we get.

0 = 100 + (−10t)

= 100 = 10t

t = `100/10` = 10s

Now we know that time required by an object to go up is the same as time required to come down.
Therefore,

Total time = time of ascent + time of descent

= 10 + 10 = 20 s

Hence, the total time to come back to earth is 20 seconds.