A ball thrown straight up takes 5.80 seconds to go up and come back down to its starting point. Neglecting air resistance, how high (in m) does it go?
41.209
It reaches its peak at time 5.80/2 seconds
So then if falls...starting at the top with zero velocity.
h=1/2 g t^2 find h, using t=2.90seconds
Well, if you throw a ball straight up, it's definitely aiming for the stars! But don't worry, it always comes back down to reality, just like my hopes and dreams. Now, to calculate how high it goes, we can use a nifty little formula:
distance = (1/2) * acceleration * time squared
Since the ball comes back down to its starting point, we can split the time into two halves. The time it takes to go up is 5.80 seconds, so we'll divide that by 2 to get 2.90 seconds.
Now, we'll plug the numbers into the formula using the acceleration due to gravity, which is approximately 9.8 m/s²:
distance = (1/2) * 9.8 m/s² * (2.90 s)²
Calculating this, we find that the ball reaches a height of approximately 42.67 meters. So, straight up it goes, and straight down it comes, just like my enthusiasm after a bad joke!
To determine the height to which the ball goes, we can use the formula for the height of an object in free fall:
h = (1/2)gt^2
Where:
h = height
g = acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
t = time
In this case, we need to find the height at the half-way point of the ball's upward and downward journey, which is half of the total time (5.80 seconds/2 = 2.90 seconds).
Plugging the values into the equation, we get:
h = (1/2)(9.8 m/s^2)(2.90 s)^2
= (1/2)(9.8 m/s^2)(8.41 s^2)
= 1/2 * 82.938 m
= 41.469 m
Therefore, the ball reaches a height of approximately 41.469 meters.