What was the x component of the initial velocity of the ball?

[wolves5]

A ball is thrown into the air from ground level. After a time t = 2 s, the ball has traveled to a position x1 = 27 m to the right of and y1 = 10 m up from where it was thrown (at this time, the x and y components of the ball's velocity are still positive). The axes show the x and y directions to be considered positive.

a) What was the x component of the initial velocity of the ball?
I got 13.5 m/s. I did (27m / 2 seconds)

b) What was the y component of the initial velocity of the ball?
Why is this not (10 m/ 2 seconds)?

c) What was the initial speed of the throw?
I don't know which equation to use. Or would you just do the square root of part a + part b?

d) What was the initial angle of the throw relative to the horizontal? Please enter your answer in degrees.
Is this tan-1 (10/27)?

e) What is the height of the ball at the top of its path?

I don't know what to do for this one.

 

[PhanthomJay]

The x comp of the velocity is correct since the velocity in the x direction is constant. In the y direction, the velocity is not constant since the ball is constantly decelertaing on its way up, so you must use one of the other kinematic equations for constant acceleration. What is the acceleration in the y direction?

 

[wolves5]

-9.8m/s2. Which equation is it?

 

[PhanthomJay]

You are given t =2 s and you know g = 9.8 m/s^2, and you are given the height y is 10 m after 2 seconds. You want to solve for the initial velocity in the y direction. Which equation would you choose?

 

[rwisz]

a) The x component of the initial velocity of the ball can be calculated using the formula v0x = (x1-x0)/t, where v0x is the initial x component of velocity, x1 is the final position in the x direction, x0 is the initial position in the x direction, and t is the time. In this case, x0 is 0 since the ball was thrown from ground level, and x1 is 27m. Thus, v0x = (27m-0m)/2s = 13.5m/s.

b) The y component of the initial velocity of the ball can be calculated using the formula v0y = (y1-y0)/t, where v0y is the initial y component of velocity, y1 is the final position in the y direction, y0 is the initial position in the y direction, and t is the time. In this case, y0 is 0 since the ball was thrown from ground level, and y1 is 10m. However, the y component of the ball's velocity is constantly changing due to the force of gravity, so it is not simply (10m/2s).

c) The initial speed of the throw can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle (in this case, the initial velocity) is equal to the sum of the squares of the other two sides (in this case, the x and y components of the initial velocity). Thus, the initial speed is given by v0 = √(v0x^2 + v0y^2) = √(13.5m/s)^2 + (v0y)^2).

d) The initial angle of the throw can be calculated using the formula θ = tan^-1(v0y/v0x), where θ is the angle and v0y and v0x are the initial y and x components of velocity, respectively. In this case, θ = tan^-1(10m/27m) ≈ 20.6°.

e) The height of the ball at the top of its path can be calculated using the formula y = y0 + v0y*t - 1/2*g*t^2, where y is the height, y0 is the initial height (0), v0y