Question

A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The ball is in the air for a total time of 8.0s. What is its initial velocity

Answer 1

Answer:

The initial velocity of the ball is 39.2 m/s in the upward direction.

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, $a=g=-9.8\ m/s^2$(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

$S=ut+\frac{1}{2}at^2$

Plug in the given values and solve for 'u'. This gives,

$0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s$

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.