Upward Moving Projectiles - determining change in time

[Kingfisher1]

I have been asked to determine the length of time the ball will be in the air. I understand the basic concept, but the execution is tripping me up. It has been years since I have done a math course, and this is my first foray into physics, so I am a bit rusty. :)

A ball is thrown upwards at a 15 degree angle. Initial velocity is 20m/s. In a previous questions V1(vertical) was determined to be 5.2 m/s. Acceleration is of course -9.81 m/s^2.

Equation used:

d= v1(t) + 1/2(a)(t)^2

d= 0
a=-9.8 m/s ^2
V1(vert)= 5.2 m/s
t= X

My efforts:

0=(5.2)t + 1/2 (-9.8)t
0= t (5.2 - 4.9)
t=0

So two issues here:
1) I am truly at a loss when re-arranging this equation, I shuffled things around but am not sure it was correct, refresh my memory please!
2) My text states that I should have two answers. I don't understand how I am to get two answers, as they do not suggest I change the equation in anyway.

Thank you for helping!

 

[Rellek]

Hey there!

You might have more help with using $$V_{f} = V_{0} - gt$$ for the y-component of your velocity.

If you find the time it will take for your ball to come to a stop in midair, by the symmetry of the arc of its trajectory, how would this time compare to the total time to be in the air? (Try drawing the trajectory to get a better idea.)

Also, for the x component of velocity, you know that there is no acceleration in this direction (because we are assuming no air resistance) so you can merely use the definition of velocity to solve for this part, $$V = \frac{x}{t}$$

 

[Kingfisher1]

Thank you for your reply! That method is much more straightforward.
I did eventually figure out what my text was trying to lead me towards with that equation.
In case it helps other beginners:
The equation properly rearranged is t=2V1/a. Therefore t= 1.1 seconds.