What Is the Initial Speed of a Ball Thrown at an Angle to Return in 3 Seconds?

[1234567890]

Homework Statement

A ball thrown straight upward returns to its original level in 3.00 s. A second ball is thrown at an angle of 36.0° above the horizontal. What is the initial speed of the second ball if it also returns to its original level in 3.00 s?
anwer : ___ m/s

Homework Equations

i need help finding the correct equation which would guide me to a correct answer

The Attempt at a Solution

 

[R.Harmon]

Consider the vertical motion of the particle. You know that v=u+at. Take the final velocity v to be zero (at the maximum height of the particle, and halfway through the motion). You know the time t (half of the total time), and the acceleration due to gravity. This let's you find the initial vertical velocity of the particle, which you can then resolve to find the initial speed. Hope that makes sense and helps!

 

[baba89]

There are a few different equations that could help solve this problem. One option is to use the equation for vertical displacement, which is given by: Δy = v0t + 1/2at^2. Since the ball returns to its original level, we know that the final displacement (Δy) is equal to 0. So, we can rearrange the equation to solve for the initial velocity (v0): v0 = -1/2at^2. We also know that the acceleration due to gravity (a) is -9.8 m/s^2, and the time (t) is 3.00 s. Plugging in these values, we get: v0 = -1/2(-9.8)(3.00)^2 = 44.1 m/s. This would be the initial velocity of the second ball if it is thrown straight up at an angle of 36.0° above the horizontal.

Another option is to use the equation for vertical velocity, which is given by: vf = v0 + at. Since the final velocity (vf) is equal to 0 when the ball reaches its original level, we can rearrange the equation to solve for the initial velocity (v0): v0 = -at. Again, we know that the acceleration due to gravity (a) is -9.8 m/s^2, and the time (t) is 3.00 s. Plugging in these values, we get: v0 = -(9.8)(3.00) = 29.4 m/s. However, this equation only gives us the vertical component of the initial velocity. To find the total initial velocity, we need to use trigonometry to find the horizontal component. We can use the angle (36.0°) and the vertical velocity (29.4 m/s) to find the horizontal velocity using the equation: v0x = v0cosθ. Plugging in the values, we get: v0x = (29.4)(cos36.0°) = 23.6 m/s. So, the total initial velocity of the second ball is 23.6 m/s at an angle of 36.0° above the horizontal.

Overall, there are multiple equations and methods that can be used to solve this problem. It is important to carefully consider the given information and choose the most appropriate equation to use.