30º above the horizontal with initial velocity problem

Nexion21

1. The problem statement, all variables and given/known data

A basketball player throws a ball at an angle of 30º above the horizontal. If the ball in has an initial velocity of 29.4 m/s, how far does it travel?

2. Relevant equations

Vyf^2 = Vyi^2 + 2(Ay)(Dy) solve for Dy and get

[Vyf^2 - (Vyi^2)] / (2(Ay)) = Dy

3. The attempt at a solution

0 - (29.4sin30)^2 / (2(-9.8)) = 11.025

29.4sin30 = 14.7 and then 14.7^2 = 216.09 and then 0 - 216.09 is -216.09/(2(-9.8)) = 11.025

CAF123

'how far does it travel' refers to a horizontal distance. First start by computing the time of flight.

Nexion21

Okay so my new attempt is Vyf = Vyi + AyT

0 = 29.4 + -9.8t

-29.4 = -9.8t

3 = t

So my time is 3, now just don't know what equation to use

CAF123

Check your value for [itex] v_{yi}. [/itex]

Nexion21

Sorry that this is going to seem hopeless but I just started taking an online course... What exactly does Vyi stand for? I thought it was the initial velocity in the vertical direction but that must be wrong

CAF123

You are correct, [itex] v_{yi} [/itex] is the vertical component of initial velocity. It is given by [itex] v_{yi} = v_i\,\sin\theta. [/itex]

Nexion21

Ooh sinθ is with it. So vyi = 14.7, which would mean t = 1.5

After that I tried using the dx = vx * t equation but that didn't work. What is the formula to solve for horizontal distance?

CAF123

What you have found is the time to the apex of the trajectory. To find the time for the whole flight, either multiply your time by 2, (valid because of symmetry of flight and constant horizontal component of velocity) or redo the whole calculation using the condition, [tex] v_{yi}\hat{y} = -v_{yf}\hat{y} [/tex]
Edit: To clarify, the vertical component of final velocity (when the ball lands back to the ground) is equal in magnitude but opposite in direction to the vertical component of initial velocity.