Question

Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the beachball's velocity at the moment it was tossed into the air?

Answer 1

Answer:

The beach ball's velocity at the moment it was tossed into the air is 4.9 m/s.

Explanation:

Given:

Time taken by the ball to reach maximum height is, $t=0.50\ s$

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, $v=0\ m/s$

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, $a=g=-9.8\ m/s^2$

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

$v=u+at$

Where, 'u' is the initial velocity.

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

$0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s$

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

Answer 2

Answer:4.9

Explanation:

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