Question

If you throw a ball up into the air and it reaches its original height after 6 seconds, calculate how high above its original position the ball reached. (Answer to two decimal places).

Answer 1

Using the equations of motion, the ball reached a height of approximately 44.14 meters above its original position after being thrown into the air and returning to its original height in 6 seconds.

Step-by-step explanation:

To determine how high above its original position the ball reached when thrown into the air and returning to its original height after 6 seconds, we can use the equations of motion for uniformly accelerated motion. Assuming acceleration due to gravity is 9.81 m/s2 (downwards) and taking upwards as the positive direction, we have the following equations:

s = ut + ½at2

Where:

t is the time.

Since the total time taken for the trip is 6 seconds, the time taken to reach the peak height is half of that, which is 3 seconds (since the time taken to go up equals the time taken to come down in symmetric projectile motion).

At the peak, the final velocity is 0, so we can use the first equation of motion:

v = u + at

0 = u - (9.81 m/s2)(3 s)

This gives us an initial velocity u of 29.43 m/s. Now we can plug this back into the displacement equation:

s = (29.43 m/s)(3 s) + ½(-9.81 m/s2)(3 s)2

This will give us the peak height above the original position:

s = 88.29 - 44.145

s = 44.145 m

The ball reached a height of approximately 44.14 meters above its original position.

Answer 2

Step-by-step explanation:

Assuming that there is no air resistance, the given time means that the ball reached its maximum height 3 seconds after throwing it upwards so we can solve its

maximum height by solving for
`v_(0y)` first:

`v_y = v_(0y) - gt \Rightarrow v_(0y) = gt`

`v_(0y) = (9.8\:\text{m/s}^2)(3\:\text{s}) = 29.4\:\text{m/s}`

Now that we know
`v_(0y),` we can now solve for
`y_(max):`

`y_(max) = v_(0y)t - (1)/(2)gt^2`

`\:\:\:\:\:\:=(29.4\:\text{m/s})(3\:\text{s}) - (1)/(2)(9.8\:\text{m/s}^2)(3\:\text{s})^2`

`\:\:\:\:\:\:=44.1\:\text{m}`