Free Fall and max altitude Question

[oconn1]

A ball is thrown vertically upward from the ground with an intial speed of 15m/s. How long does it take the ball to reach its maximum altitude? What is its maximum altitude? Determine the velocity and accelartion of the ball at t=2s.

If anyone can help me with the relatively simple problem much thanks will be given! :)

ANY HELP IS GREAT!

 

[dynamicsolo]

A ball is thrown vertically upward from the ground with an intial speed of 15m/s. How long does it take the ball to reach its maximum altitude? What is its maximum altitude? Determine the velocity and accelartion of the ball at t=2s.

One part of the questions you can answer right away is: what is the acceleration of the ball? Is it the same when the ball is thrown as it is later on, or does it change?

You are asked about the maximum altitude of the ball and when it reaches that point. The ball is traveling straight up and (presumably) falls straight back down. What is its velocity at the peak? What could you use to figure out how long it takes for that to happen? After that, you should have something else you can use to tell you how far up it gets when that happens.

 

[ogb p]

I can provide a response to the question about free fall and maximum altitude. In this scenario, the ball is thrown vertically upward from the ground with an initial speed of 15m/s. We can use the equations of motion to calculate the time it takes for the ball to reach its maximum altitude, as well as its maximum altitude, velocity, and acceleration at a specific time.

First, we need to determine the time it takes for the ball to reach its maximum altitude. We can use the equation y = y0 + v0t + (1/2)at^2, where y is the displacement, y0 is the initial displacement (in this case, 0 since the ball is thrown from the ground), v0 is the initial velocity, a is the acceleration due to gravity (9.8m/s^2), and t is the time. We know that at the maximum altitude, the velocity is 0, so we can set v = 0 and solve for t.

0 = 0 + 15t + (1/2)(-9.8)t^2

Simplifying, we get -4.9t^2 + 15t = 0

Using the quadratic formula, we get t = 0s or t = 3.06s. Since the ball cannot reach its maximum altitude in 0 seconds, we can conclude that it takes approximately 3.06 seconds for the ball to reach its maximum altitude.

Next, we can calculate the maximum altitude by plugging in t = 3.06s into the equation y = y0 + v0t + (1/2)at^2.

y = 0 + 15(3.06) + (1/2)(-9.8)(3.06)^2

Simplifying, we get y = 22.9m. Therefore, the maximum altitude of the ball is approximately 22.9 meters.

To determine the velocity and acceleration of the ball at t = 2s, we can use the equations v = v0 + at and a = a0, where a0 is the initial acceleration, which is equal to -9.8m/s^2 in this case.

Velocity at t = 2s:

v = 15 + (-9.8)(2)

v = -4.6m/s

Acceleration at t = 2s:

a =