A ball thrown horizontally with speed vi = 26.0m/s travels a horizontal distance of d = 60.0m before hitting the ground. From what height h was the ball thrown?

Which I got this answer below.
h= 26.1 m

How do I solve for the one below.

If it takes a ball dropped from rest 2.308s to fall to the ground, from what height H was it released?

Well, well, well, let's see what we've got here! We have a ball being dropped from rest and we need to find out from what height it was released. Alrighty then, let's get started!

We know that the time it takes for the ball to fall is 2.308 seconds. Now, we can use a little bit of physics magic called gravity to help us out. Gravity is a downward force that causes objects to accelerate towards the ground. In this case, we're dealing with a ball being dropped, so it's accelerating at a rate of 9.8 m/s² (approximately, for all you sticklers out there).

Now, we can use a lovely little equation: h = (1/2)gt². In this equation, h represents the height, g represents the acceleration due to gravity, and t represents the time it takes for the ball to fall. Plugging in the numbers we have, we get:

h = (1/2)(9.8 m/s²)(2.308 s)²

Calculating this little equation, we find that h ≈ 25.9 m.

And there you have it! The ball was released from a height of approximately 25.9 meters. Keep in mind that this is an approximation due to the approximate value of g. But hey, it's close enough for all the non-scientific purposes, right? Happy falling!

To solve this problem, you can use the equation of motion for free-falling objects:

H = 0.5 * g * t^2

where H is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes for the object to fall.

In this case, the time taken is 2.308 seconds. So, substituting the values into the equation:

H = 0.5 * 9.8 m/s^2 * (2.308 s)^2

H = 0.5 * 9.8 m/s^2 * 5.316864 s^2

H = 25.962416 m

Therefore, the ball was released from a height of approximately 25.96 meters (or rounding to two decimal places, 26.0 meters).

To solve this problem, we can use the kinematic equation for free fall, which relates the time of fall, the initial velocity, the acceleration due to gravity, and the height. The equation is:

h = vi * t + (1/2) * g * t^2

where:
h is the height
vi is the initial velocity (0 m/s for a ball dropped from rest)
t is the time of fall
g is the acceleration due to gravity (approximately 9.8 m/s^2)

In this case, we are given the time of fall, t = 2.308 s. We can substitute this value into the equation and solve for h.

h = 0 * 2.308 + (1/2) * 9.8 * (2.308)^2

Simplifying the equation gives:

h = (1/2) * 9.8 * 5.312064

h ≈ 25.989 m

Therefore, the ball was released from a height of approximately 25.989 meters.

a. d = Vo*t = 60 m

26 * t = 60
t = 2.31 s. = Fall time.
h = 0.5g*t^2 = 4.9*2.31^2 = 26.1 m.

b. h = Same as part a.