Solving a Rookie QB's Trajectory Problem

[Musicman]

Ok I think i am getting the hang of this stuff. But,

A rookie quaterback throws a football with an initial upward velocity component of 16.0 m/s and a horizontal velocity component of 19.0 m/s. Ignore air resistance.

a. how much time is required for the ball to reach the highes point of trajectory? Ok I did 19=1/2(9.8)t^2 to get my time which comes out to 1.96. However, it said the answer is wrong, but i used the y=1/2gt^2 because it is in the horizontal direction.

b. how high is this point?

c. how much time after it is thrown for the ball to reach its original level?
i thought this would be the same as part A

d. how far has it traveled horizontally during this time?
i know i would use d=vt, but i would need to know my time which i don't see how it is incorrect.

oh wait, should i have used 16 in place of the 19?

 

[member 553083]

For part a, you are correct in using the equation y = 1/2gt^2. However, the initial velocity for this equation should be the vertical velocity, which is 16 m/s. So, using y = 1/2(9.8)t^2 with an initial velocity of 16 m/s gives you a time of 2.04 seconds. For part b, the height of the highest point of the trajectory can be found by substituting the time from part a into the equation y = vt - 1/2gt^2. This gives you a height of 128.6 m. For part c, the time it takes for the ball to reach its original level is the same as the time to reach its highest point, which is 2.04 seconds. For part d, the distance traveled horizontally can be found by using the equation d = vt, where v is the horizontal velocity (19 m/s) and t is the time (2.04 seconds). This gives you a horizontal distance of 38.76 m.