Question

A ball is thrown up with a velocity of 20m/s.If air has no resistance,find the maximum height travelled by it and the time taken for it to return to its original position.

Answer 1

Answer:

$\tt {\pink{Given:}}$

• Initial velocity (u) = 20ms⁻¹
• Final velocity (v) = 0ms⁻¹ [The ball goes up and then stops]
• Acceleration due to gravity (g) = -10ms⁻². [We assume the value of g as negative because the ball is thrown against the gravity of earth.

$\tt{\pink{To \:\: find:}}$

• Maximum height (s).
• Time taken (t).

$\tt {\pink{Solution:}}$

Recall the three equations of motion:

$\to v=u+at\\\to s=ut+\frac{1}{2} at^2\\\to 2as=v^2-u^2$

To find the maximum height we can use the third equation.

$\Longrightarrow2as=v^2-u^2$

Substitute the values.

$\Longrightarrow 2 \times -10 \times s=0^2-20^2$

$\Longrightarrow s=\frac{-400}{-20}$

$s=20m.$

To find the time taken we can use the first equation.

$\Longrightarrow v=u+at$

Substitute the values.

$\Longrightarrow 0=20+(- 10\times t)$

$\Longrightarrow t=\frac{-20}{-10}$

$\Longrightarrow t=2 \: sec$

The sum is not over! Let us assume that the ball travelled points A and B and then it goes back to A. We just found the time taken to cover AB.

Using the fact that the air has no resistance, and according to newton's third law of motion, we assume that the time taken to travel AB will be equal to the time taken to travel BA. In other words,

$\tt {\blue{ time \: of \: ascend = time \: of \: descend}}$

Thus, the time of descend (BA) = 2 sec.

Total time = time taken from A to B +  time taken from B to A

Total time = 2sec + 2sec

Total time = 4 seconds.

$\tt {\pink{FINAL \: \: CONCLUSION:}}$

• Distance = 20m
• Time = 4 seconds.

Thank you!

HOPE THIS HELPS :D