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, [tex]t=0.50\ s[/tex]
We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, [tex]v=0\ m/s[/tex]
Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, [tex]a=g=-9.8\ m/s^2[/tex]
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:
[tex]v=u+at[/tex]
Where, 'u' is the initial velocity.
Plug in the given values and solve for 'u'. This gives,
[tex]0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s[/tex]
Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s
This question can be solved using the equations of motion.
The beachball's velocity at the moment it was tossed into the air was " 4.9 m/s".
To find out the initial velocity of the beachball, we can use the first equation of motion.
[tex]v_f = v_i + gt[/tex]
where,
vf = final velocity at highest point = 0 m/s
vi = initial velocity = ?
g = acceleration due to gravity = -9.81 m/s² (negative due to upward direction)
t = time interval = 0.5 s
Therefore,
[tex]0\ m/s = v_i + (-9.81\ m/s^2)(0.5\ s)\\\\[/tex]
vi = 4.9 m/s
Learn more about equations of motion here:
https://brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
