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?

 

[Nate810]

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.

 

[kyle8921]

I would like to provide a more accurate and detailed explanation for solving this problem. Firstly, it is important to note that in projectile motion, we can analyze the vertical and horizontal components separately. This means that the initial horizontal velocity of 19.0 m/s does not affect the vertical motion of the football.

a. To find the time required for the ball to reach the highest point of its trajectory, we need to focus on the vertical component. Using the equation y = y0 + v0y*t - 1/2gt^2, where y0 is the initial height (which we can assume to be 0 since the ball is thrown from the ground), v0y is the initial vertical velocity of 16.0 m/s, and g is the acceleration due to gravity (9.8 m/s^2), we can rearrange the equation to solve for t. This gives us t = v0y/g = 16.0/9.8 = 1.63 seconds. This is the time it takes for the ball to reach its highest point in the y direction.

b. To find the height of this point, we can use the same equation y = y0 + v0y*t - 1/2gt^2 and plug in the value of t we just found. This gives us y = 0 + 16.0*1.63 - 1/2*9.8*(1.63)^2 = 13.14 meters. So the highest point of the trajectory is 13.14 meters above the ground.

c. To find the time it takes for the ball to reach its original level, we can use the same equation with y = 0, since the ball starts and ends at the same level. This gives us 0 = 0 + 16.0*t - 1/2*9.8*t^2. Solving for t, we get t = 0 or t = 3.27 seconds. However, the time t = 0 does not make sense in this context, so we can ignore it and conclude that the ball takes 3.27 seconds to return to its original level.

d. To find the horizontal distance traveled during this time, we can use the equation x = x0 + v0x*t, where x0 is the initial horizontal position (again, we can assume it to be 0