Question

The animation shows a ball which has been kicked upward at an angle. Run the animation to watch the motion of the ball. Click initialize to set up the animation and start to run it.
Ghosts are left by the ball once per second. The animation can also be paused and moved forward in single frame mode using the step button. The cursor can be used to read the (x,y) coordinates of a position in the grid by holding down the left mouse button. Assume the grid coordinates read out in meters. When entering components, presume that x is positive to the right and y is positive upwards. Note that this ball is NOT being kicked on Earth. Do not expect an acceleration of 9.80 m/s2 downward, though you can presume that gravity is acting straight down. Use this animation to answer the following questions. Note that there are a number of different ways to go about each of the following questions. Your answer needs to be within 5% of the correct answer for credit. Please enter your answer to 3 significant digits.

What is the maximum height which the ball reaches? 42.24 m

What is the horizontal component of the initial velocity of the ball? 5.57 m/s

What is the vertical component of the initial velocity of the ball? 16.18 m/s

What is the vertical component of the acceleration of the ball? _____????

Answer 1

Answer:

The acceleration of the ball is  $a_y = - 0.3672 \ m/s^2$

Explanation:

From the question we are told that

       The maximum height the ball reachs is $H_{max} = 42.24 \ m$

       The horizontal component of the initial velocity of the ball is $v_{ix} = 5.57 \ m/s$

       The vertical component of the initial velocity of the ball is $v_{iy} = = 16.18 m/s$

The vertically motion of the ball can be mathematically represented as

       $v_{fy}^2 = v_{iy} ^2 + 2 a_{y} H_{max}$

Here the final velocity at the maximum height is zero so $v_{fy} = 0 \ m/s$

Making the acceleration $a_y$ the subject we have

        $a_y = \frac{v_{iy} ^2}{2H_{max}}$

substituting values

      $a_y = - \frac{5.57^2}{2* 42.24}$

      $a_y = - 0.3672 \ m/s^2$

The negative sign shows that the direction of the acceleration is in the negative y-axis